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

Known :

Mass (m) = 2 kg

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

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

Sin 37o = 0.6

Cos 37o = 0.8

Coefficient of the kinetic frictionk) = 0.2

The y-component of the weight (wy) = w cos 37o = (20)(0.8) = 16 Newton

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

the normal force (N) = wy = 16 Newton

Wanted : The external force (F)

Solution :

wx = 12 Newton

The force of the kinetic friction (fk) = µk N = (0.1)(16) = 1.6 Newton

The magnitude of the external force F exerted on the block :

F + fk – wx = 0

F = wx – fk

F = 12 – 1.6

F = 10.4 Newton

The external force F greater than 10.4 Newton.

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2. Mass of a block = 2 kg, coefficient of static friction µs = 0.4 and θ = 45o. Determine the magnitude of the force F so the block start to slides up.

Known :

The coefficient of the static friction (µs) = 0.4

Angle (θ) = 45o

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

Block’s mass (m) = 2 kilogram

Block’s weight (w) = m g = (2 kg)(10 m/s2) = 20 kg m/s2 = 20 Newton

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

The y-component of the weight (wy) = 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 Fwx + fs.

The x-component of the weight :

wx = 10√2 Newton

the y-component of the weight :

wy = 10√2 Newton

The normal force :

N = wy = 10√2 Newton

The force of the static friction :

fs = µs N = (0,4)(10√2) = 4√2

The magnitude of the force F so that the block starts to slide up :

Fwx + fs

F ≥ 10√2 + 4√2

F ≥ 14√2 Newton

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