Rong benda kanthi gedhene percepatan sing padha – Aplikasi hukum gerak Newton lan solusine

1. Two masses m1 = 2 kg lan 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 gesekan kinetik between m2 and incline is 0.1.

(a) Determine their percepatan

(b) Determine the tension force

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

Dikenal:

Massa 1 (m1) = 2 kg

Massa 2 (m2) = 4 kg

Coefficient of the kinetic friction between m1 lan bidang miring (μk1) = 0.2

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

Akselerasi amarga gravitasi (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 = bobot 1 = m1 g = (2 kg)(9.8 m/s)2) = 19.6 Newton

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

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

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

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

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

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

w2y = w2 kosok 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

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Gedhene percepatan:

∑Fx = max

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) lanx

w2x - Fk2 – w1x - Fk1 = (m1 +m2 ) lanx

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

Magnitudo percepatan = 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 = (4kg)(3.16m/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

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

solusi

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/s)2) = 39.2 Newton

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

a) Magnitude and direction of the acceleration

∑Fy = may

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) lany

w1 – w2 = (m1 +m2) lany

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 lan m2

aplikasi Hukum kapindho Newton on m2 :

∑Fy = may

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 = may —— ay = 0

∑Fy = 0

Upward force are positive, downward forces are negative :

T3 - T1 - T2 = 0

T3 = T1 +T2

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

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

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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

solusi

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/s)2) = 98 Newton

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

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

w2x = w2 dosa 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 = max —— 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 = may —— 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

Magnitudo gaya tarik = T1 = T2 = T = 98 Newton

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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

Solusi:

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/s)2) = 156.8 Newton

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

w1y = w1 kosok 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/s)2) = 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 lan 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 lan Fs32 are action-reaction pair)

w2 = The weight of the block 2 = m2 g = (12 kg)(9.8 m/s)2) = 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 = max

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 = max

w1x - T1 +T2 - Fk2 - Fs32 +Fs23 = (m1 +m2 +m3) lanx

w1x - Fk2 = (m1 +m2 +m3 ) lanx

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 = max

w1x - Fk2 = (m1 +m2) lanx

—–> 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

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  1. Massa lan bobot
  2. Gaya normal
  3. Hukum gerak Newton sing kapindho
  4. Gaya gesekan
  5. Gerak ing permukaan horisontal tanpa gaya gesekan
  6. Gerakan rong awak kanthi percepatan sing padha ing permukaan horisontal kasar kanthi gaya gesekan
  7. Gerak ing bidang miring tanpa gaya gesekan
  8. Gerak ing bidang miring kasar kanthi gaya gesekan
  9. Gerakan ing lift
  10. Gerakan awak disambungake dening tali lan katrol
  11. Rong benda kanthi gedhene percepatan sing padha
  12. Ngubengi kurva datar - dinamika gerakan melingkar
  13. Ngubengi kurva miring - dinamika gerakan bunder
  14. Gerakan seragam ing bunderan horisontal
  15. Gaya sentripetal ing gerakan bunder seragam

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