Rounding a flat curve – dynamics of circular motion problems and solutions

1. A 2000-kg car rounds a curve on a flat road of radius 150 m. The coefficient of static friction is 0.5. Determine the maximum speed so the car follows the curve and not skid. Acceleration due to gravity = 10 m/s2.

Known :

Mass (m) = 2000 kg

Radius (r) = 150 meters

Coefficient of static friction (μs) = 0.5

Weight (w) = m g = (2000 kg)(10 m/s2) = 20,000 kg m/s2 = 20,000 N

Force of static friction (Fs) = μs N = μs w = (0.7)(20,000 N) = 14,000 N

Wanted : v

Solution :

Rounding a flat curve – dynamics of cicular motion problems and solutions 1

[wpdm_package id=’496′]

  1. Mass and weight
  2. Normal force
  3. Newton’s second law of motion
  4. Friction force
  5. Motion on the horizontal surface without friction force
  6. The motion of two bodies with the same acceleration on the rough horizontal surface with the friction force
  7. Motion on the inclined plane without friction force
  8. Motion on the rough inclined plane with the friction force
  9. Motion in an elevator
  10. The motion of bodies connected by cord and pulley
  11. Two bodies with the same magnitude of accelerations
  12. Rounding a flat curve – dynamics of circular motion
  13. Rounding a banked curve – dynamics of circular motion
  14. Uniform motion in a horizontal circle
  15. Centripetal force in uniform circular motion

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Two bodies with the same magnitude of acceleration – Application of Newton’s law of motion problems and solutions

1. Two masses m1 = 2 kg and m2 = 5 kg are on inclined plane and are connected together by a string as shown in the figure. The coefficient of the kinetic friction between m1 and incline is 0.2 and the coefficient of the kinetic friction between m2 and incline is 0.1.

(a) Determine their acceleration

(b) Determine the tension force

Two bodies with the same magnitude of acceleration – Application of Newton's law of motion problems and solutions 1

Known :

Mass 1 (m1) = 2 kg

Mass 2 (m2) = 4 kg

Coefficient of the kinetic friction between m1 and inclined planek1) = 0.2

Coefficient of the kinetic friction between m2 and inclined plane (μk2) = 0.1

Acceleration due to gravity (g) = 9.8 m/s2

a) The magnitude and direction of the acceleration

Two bodies with the same magnitude of acceleration – Application of Newton's law of motion problems and solutions 2

w1 = weight 1 = m1 g = (2 kg)(9.8 m/s2) = 19.6 Newton

w1x = w1 sin 30o = (19.6 N)(0.5) = 9.8 Newton

w1y = w1 cos 30o = (19.6 N)(0.87) = 17 Newton

N1 = The normal force on m1 = w1y = 17 Newton

Fk1 = The force of the kinetic friction on m1 = μk1 N1 = (0.2)(17 N) = 3.4 Newton

———

w2 = weight 2 = m2 g = (4 kg)(9.8 m/s2) = 39.2 Newton

w2x = w2 sin 60o = (39.2 N)(0.87) = 34.1 Newton

w2y = w2 cos 60o = (39.2 N)(0.5) = 19.6 Newton

N2 = The normal force on m2 = w2y = 19.6 Newton

Fk2 = The force of the kinetic friction on m2 = μk2 N2 = (0.1)(19.6 N) = 1.96 Newton

———

The magnitude of the acceleration :

Fx = m ax

w2x > w1x so direction of the acceleration is the same as direction of w2x.

Forces which points along acceleration is positive and forces which has opposite direction with acceleration is negative.

w2x – Fk2 – T2 + T1 – w1x – Fk1 = (m1 + m2) ax

w2x – Fk2 – w1x – Fk1 = (m1 + m2 ) ax

34.1 N – 1.96 N – 9.8 N – 3.4 N = (2 kg + 4 kg) ax

18.94 N = (6 kg) ax

ax = 18.94 N : 6 kg

ax = 3.16 m/s2

Magnitude of the acceleration = 3.16 m/s2 . Direction of the acceleration = direction of T1 = direction of w2x

b) Magnitude of the tension force

Apply Newton’s second law on the object 2 :

w2x – Fk2 – T2 = m2 ax

34.1 N – 1.96 N – T2 = (4 kg)(3.16 m/s2)

32.14 N – T2 = 12.64 N

T2 = 32.14 N – 12.64 N = 19.5 Newton

The tension force = T = T1 = T2 = 19.5 Newton

2. m1 = 4 kg, m2 = 2 kg. Determine (a) magnitude and direction of the acceleration (b) Magnitude of the tension force which connecting m1 and m2 (c) magnitude of the tension force which connecting pulley and roof.

Two bodies with the same magnitude of acceleration – Application of Newton's law of motion problems and solutions 3

Solution

Two bodies with the same magnitude of acceleration – Application of Newton's law of motion problems and solutions 4

w1 = m1 g = (4 kg)(9.8 m/s2) = 39.2 Newton

w2 = m2 g = (2 kg)(9.8 m/s2) = 19.6 Newton

a) Magnitude and direction of the acceleration

Fy = m ay

w1 > w2 so the direction of the object is same as the direction of the weight 1 (w1). Forces which has the same direction with acceleration is positive and forces which has opposite direction with acceleration is negative.

w1 – T1 + T2 – w2 = (m1 + m2) ay

w1 – w2 = (m1 + m2) ay

39.2 N – 19.6 N = (4 kg + 2 kg) ay

19.6 N = (6 kg) ay

ay = 19.6 N : 6 kg

ay = 3.26 m/s2

Magnitude of acceleration = 3.26 m/s2. Direction of acceleration = direction of w1 .

b) Magnitude of tension force which connecting m1 and m2

Apply Newton’s second law on m2 :

Fy = m ay

w1 – T1 = m1 ay

39.2 N – T1 = (4 kg)( 3.26 m/s2)

39.2 N – T1 = 13.04 N

T1 = 39.2 N – 13.04 N

T1 = 26.16 Newton

Magnitude of the tension force which connection objects = T = T1 = T2 = 26.16 Newton

c) Magnitude of the tension force which connecting pulley and roof.

Two bodies with the same magnitude of acceleration – Application of Newton's law of motion problems and solutions 5Pulley is at rest :

Fy = m ay —— ay = 0

Fy = 0

Upward force are positive, downward forces are negative :

T3 – T1 – T2 = 0

T3 = T1 + T2

T1 and T2 have the same magnitude, T1 = T2 = T = 26.16 N :

T3 = 2T = 2(26.16 N) = 52.32 Newton

3. Block 1 (m1 = 10 kg) and block 2 (m2 = 15 kg) connected by a cord over frictionless pulley. Coefficient of the static friction between the block 2 with incline = 0.6. The coefficient of the kinetic friction between the block 2 with incline = 0.42. Determine (a) The magnitude of the minimum force F exerted on the objects so the objects accelerated upward (b) Determine the magnitude of the tension force.

Two bodies with the same magnitude of acceleration – Application of Newton's law of motion problems and solutions 6

Solution

Two bodies with the same magnitude of acceleration – Application of Newton's law of motion problems and solutions 7

w1 = The weight of the block 1 = m1 g = (10 kg)(9.8 m/s2) = 98 Newton

w2 = The weight of the block 2 = m2 g = (15 kg)(9.8 m/s2) = 147 Newton

w2y = w2 cos 30o = (147 N)(0.87) = 127.89 Newton

w2x = w2 sin 30o = (147 N)(0.5) = 73.5 Newton

N2 = The normal force on the block 2 = w2y = 127.89 Newton

Fk2 = The force of the kinetic friction on the block 2 = μk2 N2 = (0.42)(127.89 N) = 53.7 Newton

Fs2 = The force of the static friction on the block 2 = μs2 N2 = (0.6)(127.89 N) = 76.7 Newton

a) The magnitude of the minimum force F exerted on the objects so the objects accelerated upward

Fx = m ax —— ax = 0

Fx = 0

Upward forces and rightward forces are positive, downward forces and leftward forces are negative.

F – Fk2 – w2x – w1 – T2 + T1 = 0

F – Fk2 – w2x – w1 = 0

F = Fk2 + w2x + w1

F = 53.7 N + 73.5 N + 98 N

F = 225.2 Newton

b) The magnitude of the tension force

Apply Newton’s law of the motion on the block 1 :

Fy = m ay —— ay = 0

Fy = 0

T1 – w1 = 0

T1 = w1 = 98 Newton

Apply Newton’s law of the motion on the block 2 :

F – Fk2 – w2x – T2 = 0

T2 = F – Fk2 – w2x

T2 = 225.2 N – 53.7 N – 73.5 N

T2 = 98 Newton

Magnitude of the tension force = T1 = T2 = T = 98 Newton

4. Block 1 (m1 = 16 kg) lies on a horizontal surface and the block 2 (m2 = 12 kg) lies on a smooth inclined plane, connected by a cord that passes over a small, frictionless pulley. Block 3 (m3 = 5 kg) lies on the block 2. The coefficient of the kinetic friction between the block 2 and the horizontal surface is 0,4. The coefficient of the static friction between the block 2 with the block 3 is 0,3.

(a) When the system is released from rest, the block 3 and the block 2 still slide together ?

(b) If there is no block 3, what is the acceleration of the block 1 and the block 2 ?

Two bodies with the same magnitude of acceleration – Application of Newton's law of motion problems and solutions 8

Solution :

a) When the system is released from rest, the block 3 and the block 2 still slide together?

Two bodies with the same magnitude of acceleration – Application of Newton's law of motion problems and solutions 9

w1 = The weight of the block 1 = m1 g = (16 kg)(9.8 m/s2) = 156.8 Newton

w1x = w1 sin 60o = (156.8 N)(0.87) = 136.4 Newton

w1y = w1 cos 60o = (156.8 N)(0.5) = 78.4 Newton

N1 = The normal force exerted on the block 1 by the inclined plane = w1y = 78.4 Newton

w3 = The weight of the block 3 = m3 g = (5 kg)(9.8 m/s2) = 49 Newton

N23 = The normal force exerted on the block 3 bythe  block 2 = w3 = 49 Newton

N32 = The normal force exerted on the block 2 by the block 3 = N23 = w3 = 49 Newton

(N23 and N32 are action-reaction pair)

Fs23 = The force of the static friction exerted on the block 3 by the block 2 = μs N23 = (0.3)(49 N) = 14.7 Newton

Fs32 = The force of the static friction exerted on th block 2 by the block 3 = Fs23 = 14.7 Newton

(Fs23 and Fs32 are action-reaction pair)

w2 = The weight of the block 2 = m2 g = (12 kg)(9.8 m/s2) = 117.6 Newton

N2 = The normal force exerted on the object 2 by the horizontal surface = w2 + N32 = 117.6 Newton + 49

Newton = 166.6 Newton

Fk2 = The force of the kinetic friction on the block 2 = μk N2 = (0.4)(166.6 N) = 66.64 Newton

Apply Newton’s law of motion on the block 3 :

Fx = m ax

Fs23 =m3 ax

—–> Fs23 = μs N23 = μs w3 = μs m3 g

μs m3 g = m3 ax

μs g = ax

ax = (0.3)(9.8 m/s2) = 2.94 m/s2

The maximum acceleration of the block 3 so that the block 3 and the block 2 still slide together is 2.94 m/s2.

Now we calculate the magnitude of the system’s acceleration after released from rest.

The direction of the block displacement = the direction of the block’s acceleration = the direction of T2 = the direction of w1x.

Fx = m ax

w1x – T1 + T2 – Fk2 – Fs32 + Fs23 = (m1 + m2 + m3) ax

w1x – Fk2 = (m1 + m2 + m3 ) ax

136.4 N – 66.64 N = (16 kg + 12 kg + 5 kg) ax

69.76 N = (33 kg) ax

ax = 2.11 m/s2

ax is positive, means direction of the block displacement or the direction of the acceleration is same as direction of T2 or direction of w1x.

The magnitude of the acceleration is 2.11 m/s2 , lower than 2.94 m/s2 so we can conclude that block 3 and block 2 still slide together after released from rest.

b) The magnitude of the acceleration of the block 1 and the block 2

Fx = m ax

w1x – Fk2 = (m1 + m2) ax

—–> Fk2 = μk N2 = μk w2 = μk m2 g = (0.4)(12 kg)(9.8 m/s2) = 47.04 Newton

136.4 N – 47.04 N = (16 kg + 12 kg) ax

89.36 N = (28 kg) ax

ax = 89.36 N : 28 kg = 3.19 m/s2

[wpdm_package id=’493′]

  1. Mass and weight
  2. Normal force
  3. Newton’s second law of motion
  4. Friction force
  5. Motion on the horizontal surface without friction force
  6. The motion of two bodies with the same acceleration on the rough horizontal surface with the friction force
  7. Motion on the inclined plane without friction force
  8. Motion on the rough inclined plane with the friction force
  9. Motion in an elevator
  10. The motion of bodies connected by cord and pulley
  11. Two bodies with the same magnitude of accelerations
  12. Rounding a flat curve – dynamics of circular motion
  13. Rounding a banked curve – dynamics of circular motion
  14. Uniform motion in a horizontal circle
  15. Centripetal force in uniform circular motion

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Equilibrium of the bodies on a inclined plane – application of the Newton’s first law problems and solutions

1. A 2-kg block lies on a rough inclined plane at an angle 37o to the horizontal. Determine the magnitude of the external force exerted on the block, so the block is not slides down the plane. (sin 37o = 0.6, cos 37o = 0.8, g = 10 m.s-2, µk = 0.2)

Equilibrium of bodies on inclined plane – application of Newton's first law problems and solutions 1Known :

Mass (m) = 2 kg

Acceleration due to gravity (g) = 10 m/s2

Block’s weight (w) = m g = (2)(10) = 20 Newton

Sin 37o = 0.6

Cos 37o = 0.8

Coefficient of the kinetic frictionk) = 0.2

The y-component of the weight (wy) = w cos 37o = (20)(0.8) = 16 Newton

The x-component of the weight (wx) = w sin θ = (20)(sin 37) = (20)(0.6) = 12 Newton

the normal force (N) = wy = 16 Newton

Wanted : The external force (F)

Solution :

Equilibrium of bodies on inclined plane – application of Newton's first law problems and solutions 2wx = 12 Newton

The force of the kinetic friction (fk) = µk N = (0.1)(16) = 1.6 Newton

The magnitude of the external force F exerted on the block :

F + fk – wx = 0

F = wx – fk

F = 12 – 1.6

F = 10.4 Newton

The external force F greater than 10.4 Newton.

2. Mass of a block = 2 kg, coefficient of static friction µs = 0.4 and θ = 45o. Determine the magnitude of the force F so the block start to slides up.

Equilibrium of bodies on inclined plane – application of Newton's first law problems and solutions 3Known :

The coefficient of the static friction (µs) = 0.4

Angle (θ) = 45o

Acceleration due to gravity (g) = 10 m/s2

Block’s mass (m) = 2 kilogram

Block’s weight (w) = m g = (2 kg)(10 m/s2) = 20 kg m/s2 = 20 Newton

The x-component of the weight (wx) = w sin θ = (20)(sin 45) = (20)(0.5√2) = 10√2 Newton

The y-component of the weight (wy) = w cos θ = (20)(cos 45) = (20)(0.5√2) = 10√2 Newton

Wanted : The magnitude of the force F

Solution :

Equilibrium of bodies on inclined plane – application of Newton's first law problems and solutions 4Block starts to slide up, if Fwx + fs.

The x-component of the weight :

wx = 10√2 Newton

the y-component of the weight :

wy = 10√2 Newton

The normal force :

N = wy = 10√2 Newton

The force of the static friction :

fs = µs N = (0,4)(10√2) = 4√2

The magnitude of the force F so that the block starts to slide up :

Fwx + fs

F ≥ 10√2 + 4√2

F ≥ 14√2 Newton

[wpdm_package id=’492′]

  1. Particles in one-dimensional equilibrium
  2. Particles in two-dimensional equilibrium
  3. Equilibrium of bodies connected by cord and pulley
  4. Equilibrium of bodies on the inclined plane

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Equilibrium of bodies connected by cord and pulley – application of Newton’s first law problems and solutions

1. A box of mass 5 kg is on an inclined plane at an angle 30o. The box supported by a cord. Determine the tension force (T) and the normal force (N)!

Equilibrium of bodies connected by cord and pulley – application of Newton's first law problems and solutions 1

Solution

Equilibrium of bodies connected by cord and pulley – application of Newton's first law problems and solutions 2Fx = 0

T – w sin 30o = 0

T = w sin 30o

T = (5 kg)(9.8 m/s2) sin 30o

T = (49)(0.5)

T = 24.5 Newton

Fy = 0

N – w cos 30o = 0

N = w cos 30o

N = (49)(0.87)

N = 43 Newton

2. Two objects of mass m1 = m2 = 2 kg, connected by a massless string over a frictionless pulley. Find the tension force T1 and T2.

Equilibrium of bodies connected by cord and pulley – application of Newton's first law problems and solutions 3

Solution

Equilibrium of bodies connected by cord and pulley – application of Newton's first law problems and solutions 4

(a) Free-body diagram for object 1 (b) Free-body diagram for object 2

Apply Newton’s first law to object 1 :

Fy = 0

T1 – w1 = 0

T1 = w1 = m1 g = (2 kg)(9.8 m/s2) = 19.6 N

Apply Newton’s first law to object 2 :

Fy = 0

T2 – w2 = 0

T2 = w2 = m2 g = (2 kg)(9.8 m/s2) = 19.6 N

T1 = T2 = 19.6 N.

3. An object of weight wA = 30 N and an object of weight wB = 40 N, are attached by a lightweight cord that passes over a frictionless pulley of the negligible mass. Determine the coefficient of the maximum static friction between wB and inclined surface, if the system is at rest.

Equilibrium of bodies connected by cord and pulley – application of Newton's first law problems and solutions 5

Solution

Equilibrium of bodies connected by cord and pulley – application of Newton's first law problems and solutions 6

(a) Free-body diagram for object wA (b) Free-body diagram for object wB

Apply Newton’s first law to object wA in vertical (y) direction :

Fy = 0 (no acceleration in vertical direction)

T – wA = 0

T = wA = 30 Newton

Apply Newton’s first law to object wB in vertical (y) direction :

Fy = 0

N – wB cos 45o = 0

N = wB cos 45o = (40)(0.7) = 28 Newton

Apply Newton’s first law to object wB in horizontal (x) direction :

Fx = 0

Fk + wB sin 45o – T = 0

μs N + wB sin 45o – T = 0

μs (28) + (40)(0.7) – 30 = 0

μs (28) + 28 – 30 = 0

μs (28) = 30 – 28

μs (28) = 2

μs = 2 / 28

μs = 0.07

The coefficient of the maximum static friction between wB and inclined surface = 0.07.

[wpdm_package id=’490′]

  1. Particles in one-dimensional equilibrium
  2. Particles in two-dimensional equilibrium
  3. Equilibrium of bodies connected by cord and pulley
  4. Equilibrium of bodies on inclined plane

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Particles in two-dimensional equilibrium – application of Newton’s first law problems and solutions

1. Find the tension force T1, T2, and T3. Ignore cord’s mass.

Particles in two-dimensional equilibrium – application of Newton's first law problems and solutions 1

Solution

Particles in two-dimensional equilibrium – application of Newton's first law problems and solutions 2

(a) Free-body diagram for object (b) Free-body diagram for cord

Apply the Newton’s first law on the object :

ΣFy = 0

T1 – w = 0

T1 = w = m g

T1 = (5 kg)(9.8 m/s2)

T1 = 49 kg m/s2

T1 = 49 N

Apply Newton’s first law on the cord :

Fx = 0

T3x – T 2x = 0

T3 cos 30o – T2 cos 40o = 0

0.87 T3 – 0.77 T2 = 0

0.87 T3 = 0.77 T2

T2 = 0.87 T3 / 0.77 = 1.1 T3 ———- Equation 1

——

Fy = 0

T3y + T2y – T1y = 0

T3 sin 30o + T2 sin 40o – T1 = 0

0.5 T3 + 0.64 T2 – 49 N = 0 ———- Equation 2

Substituting T2 in the equation 2 into the equation 2 :

0.5 T3 + 0.64 (1.1 T3) – 49 N = 0

0.5 T3 + 0.70 T3 – 49 = 0

1.2 T3 – 49 = 0

1.2 T3 = 49

T3 = 49 / 1.2

T3 = 41 N

———

T2 = 1.1 T3

T2 = (1.1)(40.8 N)

T2 = 45 N

[wpdm_package id=’488′]

  1. Particles in one-dimensional equilibrium
  2. Particles in two-dimensional equilibrium
  3. Equilibrium of bodies connected by cord and pulley
  4. Equilibrium of bodies on inclined plane

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Particles in the one-dimensional equilibrium – application of the Newton’s first law problems and solutions

1. Mass of an object, m = 10 kg, supported by a cord. Find the tension in the cord! g = 10 m/s2

Particles in one-dimensional equilibrium – application of Newton's first law problems and solutions 1Known :

Mass (m) = 10 kg

Acceleration due to gravity (g) = 10 m/s2

Wanted : The tension force (T)

Solution :

ΣFy = 0

T – w = 0

T = w

T = m g

T = (10 kg)(10 m/s2) = 100 kg m/s2

T = 100 Newton

2. Mass of the object is 10 kg. Find the tension in the cord….. Acceleration due to gravity = 10 m/s2.

Solution

Known :

Mass (m) = 10 kg

Acceleration due to gravity (g) = 10 m/s2.

Wanted : The tension force (T)

Solution :

Particles in one-dimensional equilibrium – application of Newton's first law problems and solutions 2w = weight = m g = (10 kg)(10 m/s2) = 100 kg m/s2

T1 = the tension force 1

T1x = the x-component of the tension force 1 = T1 cos 45o = 0.7 T1

T1y = the y-component of the tension force 2 = T1 sin 45o = 0.7 T1

T2 = the tension force 2

T2x = the x-component of the tension force 2 = T2 cos 45o = 0.7 T2

T2y = the y-component of the tension force 2 = T2 sin 45o = 0.7 T2

The equilibrium condition ΣF = 0.

y axis :

ΣFy = 0

T1y + T2y – w = 0

0.7T1 + 0.7T2 – 100 = 0

0.7T1 + 0.7T2 = 100 —– equation 1

x axis :

ΣFx = 0

T2x – T1x = 0

0.7T2 – 0.7T1 = 0

0.7T2 = 0.7T1

T2 = T1 —– equation 2

Determine magnitude of T1 :

0.7T1 + 0.7T1 = 100

1.4T1 = 100

T1 = 100 / 1.4

T1 = 71.4 Newton

T1 = T2 so T2 = 71.4 Newton

[wpdm_package id=’486′]

  1. Particles in one-dimensional equilibrium
  2. Particles in two-dimensional equilibrium
  3. Equilibrium of bodies connected by cord and pulley
  4. Equilibrium of bodies on inclined plane

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Bodies connected by the cord and pulley – application of Newton’s law of motion problems and solutions

1. Two boxes are connected by a cord running over a pulley. Ignore the mass of the cord and pulley and any friction in the pulley. Mass of the box 1 = 2 kg, mass of the box 2 = 3 kg, acceleration due to gravity = 10 m/s2. Find (a) The acceleration of the system (b) The tension in the cord!

Bodies connected by cord and pulley - application of Newton's law of motion problems and solutions 1

Solution

Bodies connected by cord and pulley - application of Newton's law of motion problems and solutions 2Known :

Mass of the box 1 (m1) = 2 kg

Mass of the box 2 (m2) = 3 kg

Acceleration due to gravity (g) = 10 m/s2

Weight of the box 1 (w1) = m1 g = (2)(10) = 20 Newton

Weight of the box 2 (w2) = m2 g = (3)(10) = 30 Newton

Solution :

(a) magnitude and direction of the acceleration

w2 > w1 so the box 2 accelerates downward and the box 1 accelerates upward.

Forces that has the same direction with acceleration (w2 and T1), its sign positive. Forces that has opposite direction with acceleration (T2 and w1), its sign negative.

F = m a

w2 – T2 + T1 – w1 = (m1 + m2) a ——-> T1 = T2 = T

w2 – T + T – w1 = (m1 + m2) a

w2 – w1 = (m1 + m2) a

30 – 20 = (2 + 3) a

10 = 5 a

a = 10 / 5

a = 2 m/s2

Magnitude of the acceleration is 2 m/s2.

(b) The tension force

The box 2 :

There are two forces acts on the box 2 : first, weight of the box 2 (w2), points downward so it’s positive. Second, tension force exerted on the box 2 (T2), points upward so it’s negative. Apply Newton’s second law of motion.

F = m a

w2 – T2 = m2 a

30 – T2 = (3)(2)

30 – T2 = 6

T2 = 30 – 6

T2 = 24 Newton

Box 1 :

There are two forces acts on the box 1. First, weight of the box 1 (w1), points downward so it’s negative. Second, the tension force exerted on the box 1 (T1) points upward so it’s positive. Apply Newton’s second law of motion :

F = m a

T1 – w1 = m1 a

T1 – 20 = (2)(2)

T1 – 20 = 4

T1 = 20 + 4

T1 = 24 Newton

Magnitude of the tension force = T1 = T2 = T = 24 Newton

2. An object on a rough horizontal surface. Mass of the object 1 = 2 kg, mass of the object 2 = 4 kg, acceleration due to gravity = 10 m/s2, coefficient of the static friction = 0.4, coefficient of the kinetic friction = 0.3. The system is at rest or accelerated ? If the system is accelerated, find the magnitude and direction of the system’s acceleration!

Bodies connected by cord and pulley - application of Newton's law of motion problems and solutions 3

Solution

Bodies connected by cord and pulley - application of Newton's law of motion problems and solutions 4Known :

Mass of the object 1 (m1) = 2 kg

Mass of the object 2 (m2) = 4 kg

Acceleration due to gravity (g) = 10 m/s2

Coefficient of the static friction (μs) = 0.4

The coefficient of the kinetic friction (μk) = 0.3

Weight of the object 1 (w1) = m1 g = (2)(10) = 20 Newton

Weight of the object 2 (w2) = m2 g = (4)(10) = 40 Newton

Normal force exerted on the object 1 (N) = w1 = 20 Newton

Force of the static friction exerted on the object 1 (fs) = μs N = (0.4)(20) = 8 Newton

Force of the kinetic friction exerted on the object 1 (fk) = μk N = (0.3)(20) = 6 Newton

Wanted: acceleration (a)

Solution :

w2 > fs (40 Newton > 8 Newton) so the object 2 is accelerated vertically downward and the object 1 is accelerated horizontally rightward. The friction force that acts on the objects 1 is the force of the kinetic friction (fk). Apply Newton’s second law of motion :

F = m a

w2 – the = (m1 + m2) a

40 – 6 = (2 + 4) a

34 = 6 a

a = 34 / 6 = 17 / 3

a = 5.7 m/s2

Magnitude of the acceleration = 5.7 m/s2

[wpdm_package id=’484′]

  1. Mass and weight
  2. Normal force
  3. Newton’s second law of motion
  4. Friction force
  5. Motion on horizontal surface without friction force
  6. The motion of two bodies with the same acceleration on the rough horizontal surface with the friction force
  7. Motion on the inclined plane without friction force
  8. Motion on the rough inclined plane with the friction force
  9. Motion in an elevator
  10. The motion of bodies connected by cord and pulley
  11. Two bodies with the same magnitude of accelerations
  12. Rounding a flat curve – dynamics of circular motion
  13. Rounding a banked curve – dynamics of circular motion
  14. Uniform motion in a horizontal circle
  15. Centripetal force in uniform circular motion

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Application of the Newton’s law of the motion in an elevator – problems and solutions

1. A 50-kg person in an elevator. Acceleration due to gravity = 10 m/s2. Determine the normal force exerted on the object by the elevator, if :

(a) the elevator is at rest

(b) the elevator is moving downward at a constant velocity

(c) elevator accelerated upward at a constant acceleration 5 /s2

(d) elevator accelerated downward at a constant 5 m/s2

(e) elevator in a free fall

Solution

Application of Newton's law of motion on elevator - problems and solutions 1Known :

Person’s mass (m) = 50 kg

Acceleration due to gravity (g) = 10 m/s2

Weight (w) = m g = (50)(10) = 500 Newton

Wanted: The normal force (N)

Solution :

(a) the elevator is at rest

The elevator is at rest so there is no acceleration (a = 0)

We choose the upward direction in the positive direction and the downward direction in the negative direction.

ΣF = m a

N – w = 0

N = w

N = 500 Newton

(b) the elevator is moving downward at a constant velocity

Constant velocity so there is no acceleration (a = 0)

We choose the upward direction in the positive direction and the downward direction in the negative direction.

ΣF = m a

N – w = 0

N = w

N = 500 Newton

(c) elevator accelerated upward at a constant 5 m/s2

The direction of the acceleration is upward, so we choose the positive direction as up.

N – w = m a

N = w + m a

N = 500 + (50)(5)

N = 500 + 250

N = 750 Newton

The person feels the floor pushing up harder than when the elevator is stationary or moving with a constant velocity.

If the person stands on a scale, the scale reads the magnitude of the downward force exerted by the person on the scale. By Newton’s third law, this equals the magnitude of the upward normal force exerted by the scale on the person.

(d) elevator accelerated downward at a constant 5 m/s2

The direction of the acceleration is downward, so we choose the positive direction as down.

w – N = m a

N = w – m a

N = 500 – (50)(5)

N = 500 – 250

N = 250 Newton

The person’s weight is 250 N, less than actual weight w = 500 N.

(e) elevator in a free fall

Free fall means the elevator’s acceleration is the same as the acceleration due to gravity. The magnitude of the acceleration due to gravity is 9,8 m/s2, it’s direction is downward toward the center of the Earth. The speed increases linearly in time by 9,8 m/s during each second.

The direction of the acceleration is downward, so we choose the positive direction as down.

w – N = m a

N = w – m a

N = 500 – (50)(10)

N = 500 – 500

N = 0

2. Determine tension in an elevator cable. Elevator’s mass = 2000 kg.

(a) elevator is at rest

(b) elevator accelerated downward at a constant 5 m/s2

(c) elevator accelerated upward at a constant 5 m/s2

(d) elevator in a free fall

Acceleration due to gravity (g) = 10 m/s2

Solution

Application of Newton's law of motion on elevator - problems and solutions 2Known :

Elevator’s mass (m) = 2000 kg

Acceleration of gravity (g) = 10 m/s2

weight (w) = m g = (2000)(10) = 20,000 Newton

Wanted : The tension force (T)

Solution :

(a) elevator is at rest

elevator is at rest so there is no acceleration (a = 0)

We choose the upward direction as the positive direction and the downward direction as the negative direction.

ΣF = m a

T – w = 0

T = w

T = 20,000 Newton

Tension in cable (T) = elevator’s weight (w) = 20,000 Newton

(b) elevator accelerated downward at a constant 5 m/s2

The direction of the acceleration is downward, so we choose the positive direction as down.

w – T = m a

T = w – m a

T = 20,000 – (2000)(5)

T = 20,000 – 10,000

T = 10,000 Newton

c) elevator accelerated upward at a constant 5 m/s2

The direction of the acceleration is downward, so we choose the positive direction as up.

T – w = m a

T = w + m a

T = 20,000 + (2000)(5)

T = 20,000 + 10,000

T = 30,000 Newton

(d) elevator in a free fall

The direction of the acceleration is downward, so we choose the positive direction as down.

w – T = m a

T = w – m a

T = 20,000 – (2000)(10)

T = 20,000 – 20,000

T = 0

[wpdm_package id=’482′]

  1. Mass and weight
  2. Normal force
  3. Newton’s second law of motion
  4. Friction force
  5. Motion on the horizontal surface without friction force
  6. The motion of two bodies with the same acceleration on rough horizontal surface with friction force
  7. Motion on inclined plane without friction force
  8. Motion on the rough inclined plane with the friction force
  9. Motion in an elevator
  10. The motion of bodies connected by cord and pulley
  11. Two bodies with the same magnitude of accelerations
  12. Rounding a flat curve – dynamics of circular motion
  13. Rounding a banked curve – dynamics of circular motion
  14. Uniform motion in a horizontal circle
  15. Centripetal force in uniform circular motion

Read more

Motion on the rough inclined plane with the friction force – application of Newton’s law of motion problems and solutions

1. Object’s mass = 2 kg, acceleration due to gravity = 9.8 m/s2, coefficient of the static friction = 0.2, coefficient of the kinetic friction = 0.1. Is the object at rest or accelerating? If the object is accelerated, find (a) the net force (b) magnitude and direction of the box’s acceleration!

Motion on rough incline plane with friction force - application of Newton's law of motion problems and solutions 1

Solution

Motion on rough incline plane with friction force - application of Newton's law of motion problems and solutions 2

Known :

Mass (m) = 2 kg

Acceleration due to gravity (g) = 9.8 m/s2

Coefficient of the static friction (μs) = 0.2

Coefficient of the kinetic friction (μk) = 0.1

Weight (w) = m g = (2)(9.8) = 19.6 Newton

The horizontal component of the weight (wx) = w sin 30o = (19.6)(0.5) = 9.8 Newton

The vertical component of th weight (wy) = w cos 30o = (19.6)(0.5√3) = 9.8√3 Newton

The normal force (N) = wy = 9.8√3 Newton

Force of the static friction (fs) = (0.2)(9.8√3) = 1.96√3 Newton = 3.39 Newton

Force of the kinetic friction (fk) = (0.1)(9.8√3) = 0.98√3 Newton = 1.69 Newton

Solution :

Object is at rest if wx < fs, object is moving down if wx > fs.

wx = 9.8 Newton and fs = 3.39 Newton.

(a) the net force

F = wx – fk = 9.8 – 1.69 = 8.11 Newton

(b) magnitude and direction of the acceleration

F = m a

8.11 = (2) a

a = 4.05

Magnitude of the acceleration = 4.05 m/s2 and direction of the acceleration = downward.

2. Object’s mass = 4 kg, acceleration due to gravity = 9,8 m/s2. Coefficient of the kinetic friction = 0.2 and coefficient of the static friction = 0.4. Magnitude of the force F = 40 Newton. The object is at rest or slides down ? If the object slides down, find (a) the net force (b) magnitude and direction of the acceleration!

Motion on rough incline plane with friction force - application of Newton's law of motion problems and solutions 3

Solution

Motion on rough incline plane with friction force - application of Newton's law of motion problems and solutions 4

Known :

Mass (m) = 4 kg

Acceleration due to gravity (g) = 9.8 m/s2

The coefficient of the static friction (μs) = 0.4

The coefficient of the kinetic friction (μk) = 0.2

Weight (w) = m g = (4)(9.8) = 39.2 Newton

The horizontal component of the weight (wx) = w sin 30o = (39.2)(0.5) = 19.6 Newton

The vertical component of the weight (wy) = w cos 30o = (392)(0..5√3) = 19.6√3 Newton

The normal force (N) = wy = 19.6√3 Newton = 33.95 Newton

the static friction force (fs) = μs N = (0,4)(33.95) = 13.58 Newton

The kinetic friction force (fk) = μk N = (0.2)(33.95) = 6.79 Newton

F = 40 Newton

Solution :

The object slides down if F < wx + fs. The object slides up if F > wx + fs.

F = 40 Newton, wx = 19.6 Newton and fs = 13.58 Newton.

F is greater than wx + fs so the object slides up.

(a) The net force

F = F – wxfk = 40 – 19.6 – 6.79 = 13.61 Newton

(b) The magnitude and direction of the acceleration

F = m a

6.4 = (4) a

a = 1.6

The magnitude of the acceleration is 1.6 m/s2 and direction of the acceleration is upward.

[wpdm_package id=’481′]

  1. Mass and weight
  2. Normal force
  3. Newton’s second law of motion
  4. Friction force
  5. Motion on the horizontal surface without friction force
  6. The motion of two bodies with the same acceleration on the rough horizontal surface with the friction force
  7. Motion on the inclined plane without friction force
  8. Motion on the rough inclined plane with the friction force
  9. Motion in an elevator
  10. The motion of bodies connected by cord and pulley
  11. Two bodies with the same magnitude of accelerations
  12. Rounding a flat curve – dynamics of circular motion
  13. Rounding a banked curve – dynamics of circular motion
  14. Uniform motion in a horizontal circle
  15. Centripetal force in uniform circular motion

Read more

Motion on the inclined plane without the friction force – application of Newton’s law of motion problems and solutions

1. Box’s mass = 2 kg, acceleration due to gravity = 9.8 m/s2. Find (a) the net force which accelerates the box downward (b) magnitude of the box’s acceleration.

Motion on incline plane without friction force - application of Newton's law of motion problems and solutions 1

Solution

Motion on incline plane without friction force - application of Newton's law of motion problems and solutions 2

Known :

Mass (m) = 2 kg

Acceleration due to gravity (g) = 9.8 m/s2

Weight (w) = m g = (2)(9.8) = 19.6 Newton

wx = w sin 30 = (19.6)(0.5) = 9.8 Newton

wy = w cos 30 = (19.6)(0.5√3) = 9.8√3 Newton

Solution :

(a) The net force which accelerates the box

Inclined plane is smooth, so there is no friction force. The only force which acts on the object is wx.

F = wx

F = 9.8 Newton

(b) magnitude of the acceleration

F = m a

9.8 = (2) a

a = 9.8 / 2

a = 4.9 m/s2

Magnitude of the acceleration is 4.9 m/s2, direction of the acceleration is downward.

2. Inclined plane is smooth so there is no friction force. Object’s mass is 3 kg, acceleration due to gravity is 9.8 m/s2. Determine the magnitude of the force F if (a) object is at rest (b) object is moving downward with constant acceleration 2 m/s2 (c) object is moving upward with a constant acceleration of 2 m/s2.

Motion on incline plane without friction force - application of Newton's law of motion problems and solutions 3

Solution

Motion on incline plane without friction force - application of Newton's law of motion problems and solutions 4

Known :

Mass (m) = 3 kg

Acceleration due to gravity (g) = 9.8 m/s2

Weight (w) = m g = (3)(9.8) = 29.4 Newton

wx = w sin 30 = (29.4)(0.5) = 14.7 Newton

wy = w cos 30 = (29.4)(0.5√3) = 14.7√3 Newton

Solution :

(a) The magnitude of the force F if an object is at rest

Newton’s first law of motion states that if an object is at rest, the net force acts on the object is zero.

F = 0

F – wx = 0

F = wx

F = 14.7 Newton

(b) The magnitude of the force F if an object is moving downward at a constant 2 m/s2

F = m a

wx – F = m a

14.7 – F = (3)(2)

14.7 – F = 6

F = 14.7– 6

F = 8.7 Newton

(c) The magnitude of the force F if an object is moving upward at a constant 2 m/s2

F = m a

F – wx = m a

F – 14.7 = (3)(2)

F – 14.7 = 6

F = 14.7 + 6

F = 20.7 Newton

[wpdm_package id=’479′]

  1. Mass and weight
  2. Normal force
  3. Newton’s second law of motion
  4. Friction force
  5. Motion on the horizontal surface without friction force
  6. The motion of two bodies with the same acceleration on the rough horizontal surface with the friction force
  7. Motion on the inclined plane without friction force
  8. Motion on the rough inclined plane with the friction force
  9. Motion in an elevator
  10. The motion of bodies connected by cord and pulley
  11. Two bodies with the same magnitude of accelerations
  12. Rounding a flat curve – dynamics of circular motion
  13. Rounding a banked curve – dynamics of circular motion
  14. Uniform motion in a horizontal circle
  15. Centripetal force in uniform circular motion

Read more