# 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 wx. Inclined plane is smooth so there is no friction force.

wx = w sin θ = (w)(y / z)

wx = (50)(4/5)

wx = 40 Newton

Acceleration of block (a) ?

ΣF = m a

wx = m a

40 = (5) a

a = 40 / 5

a = 8 m/s2

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

Known : Coefficient of static friction (µs) = 0.5

Angle (θ) = 45o

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 Fwx + fs. Horizontal component of weight :

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

Vertical component of weight :

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

N = wy = 40√2

Force of static friction :

fs = µs N = (0.5)(40√2) = 20√2

Magnitude of force F :

Fwx + fs

F ≥ 40√2 + 20√2

F ≥ 60√2 Newton

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

Cos 37o = 0.8

Coefficient of kinetic friction (µk) = 0.1

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

Wanted : Magnitude of force F

Solution : The block is not slides down if F = wx

Horizontal component of weight parallel with inclined plane surface :

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

fk = µk N = (0.1)(64) = 6.4

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

F + fk – wx = 0

F = wx – fk = 48 – 6.4 = 41.6 Newton

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4. A 5-kg block is pulled along rough inclined plane by a force of 71 N (g = 10 m.s-2, sin 37o = 0.6, cos 37o = 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 37o = 0.6

Cos 37o = 0.8

Coefficient of friction = 0.4

Wanted : Acceleration of the block (a)

Solution :

Horizontal component of weight :

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

Vertical component of weight :

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

Normal force : N = wy = 40

Force of kinetic friction :

fk = µk N = (0.4)(40) = 16

Net force :

F = F – wx – fk = 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 (wx).

wx = w sin θ = (500 N)(0.5) = 250 N 