# 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 37^{o }to the horizontal. Determine the magnitude of the external force exerted on the block, so the block is not slides down the plane. (sin 37^{o} = 0.6, cos 37^{o} = 0.8, g = 10 m.s^{-2}, µ_{k} = 0.2)

__Known :__

Mass (m) = 2 kg

Acceleration due to gravity (g) = 10 m/s^{2}

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

Sin 37^{o} = 0.6

Cos 37^{o} = 0.8

Coefficient of the kinetic friction (µ_{k}) = 0.2

The y-component of the weight (w_{y}) = w cos 37^{o }= (20)(0.8) = 16 Newton

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

the normal force (N) = w_{y }= 16 Newton

__Wanted __ : The external force (F)

__Solution __:

w_{x} = 12 Newton

The force of the kinetic friction (f_{k}) = µ_{k} N = (0.1)(16) = 1.6 Newton

__The magnitude of the external force F exerted on the block__ :

F + f_{k} – w_{x }= 0

F = w_{x }– f_{k}

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 µ_{s }= 0.4 and θ = 45^{o}. Determine the magnitude of the force F so the block start to slides up.

__Known :__

The coefficient of the static friction (µ_{s}) = 0.4

Angle (θ) = 45^{o}

Acceleration due to gravity (g) = 10 m/s^{2}

Block’s mass (m) = 2 kilogram

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

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

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

__Wanted__ : The magnitude of the force F

__Solution :__

Block starts to slide up, if *F* ≥ *w*_{x} +* f*_{s}.

The x-component of the weight :

w_{x} = 10√2 Newton

__the y-component of the weight __:

w_{y} = 10√2 Newton

__The normal force __:

N = w_{y} = 10√2 Newton

__The force of the static friction__ :

*f*_{s} = µ_{s }N = (0,4)(10√2) = 4√2

__The magnitude of the force F so that the block starts to slide up__ :

*F* ≥ *w*_{x} +* f*_{s}

*F* ≥ 10√2 +* 4*√2

*F* ≥ 14√2 Newton

### Ebook PDF equilibrium of bodies on inclined plane sample problems with solutions

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- Equilibrium of bodies connected by cord and pulley
- Equilibrium of bodies on the inclined plane