Engineering Mechanics

(^134) „„„„„ A Textbook of Engineering Mechanics
Solution. Given: Weight of the body (W) = 400 N and coefficient of friction (μ) = 0.3.
Whether it is necessary to push the body down the plane or hold it back from sliding down.
We know that
1.2
tan 0.5 or 26.5
2.4
α= = α= °
and normal reaction, R = W cos α = 400 cos 26.5° N
= 400 × 0.8949 = 357.9 N
∴ Force of friction,
F = μR = 0.3 × 357.9 = 107.3 N ...(i)
Now resolving the 400 N force along the plane
= 400 sin α = 400 × sin 26.5° N
= 400 × 0.4462 = 178.5 N ...(ii)
We know that as the force along the plane (which is responsible for sliding the body) is more
than the force of friction, therefore the body will slide down. Or in other words, it is not necessary to
push the body down the plane, rather it is necessary to hold it back from sliding down. Ans.
Minimum force required parallel to the plane
We know that the minimum force required parallel to the plane to hold the body back,
P = 178.5 – 107.3 = 71.2 N Ans.
Example 8.7. An effort of 200 N is required just to move a certain body up an inclined
plane of angle 15° the force acting parallel to the plane. If the angle of inclination of the plane is
made 20° the effort required, again applied parallel to the plane, is found to be 230 N. Find the
weight of the body and the coefficient of friction.
Solution. Given: First case : When effort (P 1 ) = 200 N, then angle of inclination (α 1 ) = 15°
and second case : When effort (P 2 ) = 230 N, then angle of inclination (α 2 ) = 20°.
Let μ = Coefficient of friction,
W = Weight of the body,
R = Normal reaction, and
F = Force of friction.
Fig. 8.12.
First of all, consider the body lying on a plane inclined at an angle of 15° with the horizontal
and subjected to an effort of 200 N as shown in Fig. 8.12 (a).
Resolving the forces at right angles to the plane,
R 1 = W cos 15° ...(i)
Fig. 8.11.