Inclined plane – problems and solutions

1. Mass of block is 5 kg, acceleration due to gravity = 9.8 m/s², tan 37 ^o = 3/4. What is the acceleration of the block?

Solution

Known :

mass (m) = 5 kg

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

weight (w) = m g = (5)(10) = 50 N

x = 3

The height of inclined plane (y) = 4

The length of inclined plane surface (l) = 5

Wanted : acceleration of block (a) ?

Solution :

The block is accelerated by force of w_x. Inclined plane is smooth so there is no friction force.

w_x = w sin θ = (w)(y / z)

w_x = (50)(4/5)

w_x = 40 Newton

Acceleration of block (a) ?

ΣF = m a

w_x = m a

40 = (5) a

a = 40 / 5

a = 8 m/s²

[irp]

2. A block initially at rest, then accelerated upward parallel with the inclined plane surface. Block’s mass = 8 kg, coefficient of static friction µ_s = 0.5 and angle θ = 45^o. What is the magnitude of force F so that the block is accelerated.

Known :

Coefficient of static friction (µ_s) = 0.5

Angle (θ) = 45^o

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

Block’s mass (m) = 8 kg

Block weight (w) = m g = (8 kg)(10 m/s²) = 80 kg m/s² = 80 Newton

Wanted : Magnitude of force F

Solution :

Block start to accelerated if F ≥ w_x + f_s.

Horizontal component of weight :

w_x = w sin θ = (80)(sin 45) = (80)(0.5√2) = 40√2

Vertical component of weight :

w_y = w cos θ = (80)(cos 45) = (80)(0.5√2) = 40√2

Normal force :

N = w_y = 40√2

Force of static friction :

f_s = µ_s N = (0.5)(40√2) = 20√2

Magnitude of force F :

F ≥ w_x + f_s

F ≥ 40√2 + 20√2

F ≥ 60√2 Newton

[irp]

3. Mass of block is 8 kg. The minimal force required to hold the block so that the block not slides down is …(sin 37^o = 0.6, cos 37^o = 0.8, g = 10 m.s^-2, µ_k = 0.1)

Known :

Mass of block (m) = 8 kg

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

Block’s weight (w) = m g = (8)(10) = 80 Newton

Sin 37^o = 0.6

Cos 37^o = 0.8

Coefficient of kinetic friction (µ_k) = 0.1

Normal force (N) = w_y = w cos 37^o = (80)(0.8) = 64 Newton

Wanted : Magnitude of force F

Solution :

The block is not slides down if F = w_x

Horizontal component of weight parallel with inclined plane surface :

w_x = w sin θ = (80)(sin 37) = (80)(0.6) = 48

f_k = µ_k N = (0.1)(64) = 6.4

The minimum of force F so that the block is not slides down :

F + f_k – w_x = 0

F = w_x – f_k = 48 – 6.4 = 41.6 Newton

[irp]

4. A 5-kg block is pulled along rough inclined plane by a force of 71 N (g = 10 m.s^-2, sin 37^o = 0.6, cos 37^o = 0.8). If the coefficient of friction force between block and inclined plane is 0.4, what is the acceleration of the block.

Known :

Object’s mass (m) = 5 kg

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

weight of block (w) = m g = (5)(10) = 50 Newton

Force F = 71 Newton

Sin 37^o = 0.6

Cos 37^o = 0.8

Coefficient of friction = 0.4

Wanted : Acceleration of the block (a)

Solution :

Horizontal component of weight :

w_x = w sin θ = (50)(sin 37) = (50)(0.6) = 30

Vertical component of weight :

w_y = w cos θ = (50)(cos 37) = (50)(0.8) = 40

Normal force :

N = w_y = 40

Force of kinetic friction :

f_k = µ_k N = (0.4)(40) = 16

Net force :

∑F = F – w_x – f_k = 71 – 30 – 16 = 25 Newton

Acceleration of the object :

a = ∑F / m = 25 / 5 = 5 m/s²

5. A 500-N box raised to truck through inclined plane, as shown in figure below. If height of truck is 1.5 meters, determine the magnitude of force to moving the box.

Known :

weight of box (w) = 500 Newton

height of truck (h) = 1.5 meters

length of inclined plane (l) = 3 meters

Wanted : Magnitude of force to moving the box (F)

Solution :

Sin θ = opp / hyp = 1.5 meters / 3 meters = 1.5 / 3 = 0.5

Force (F) required to moving the box, same as the horizontal component of weight which is parallel to inclined plane (w_x).

w_x = w sin θ = (500 N)(0.5) = 250 N