# Inclined plane – problems and solutions

1. Mass of block is 5 kg, acceleration due to gravity = 9.8 m/s^{2}, 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/s^{2}

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/s^{2}

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/s^{2}

Block’s mass (m) = 8 kg

Block weight (w) = m g = (8 kg)(10 m/s^{2}) = 80 kg m/s^{2} = 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

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

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/s^{2}

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

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/s^{2}

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/s^{2}

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