Question

A body of weight 500 N is pulled up along an inclined plane having an inclination of 30° with the horizontal. If the coefficient of friction between the body and the plane is 0.3 and the force is applied parallel to inclined plane, determine the force required.​

Answer 1

Answer:

body of weight 500 N is pulled up along an inclined plane having an inclination of 30° with the horizontal. If the coefficient of friction between the body and the plane is 0.3 and the force is applied parallel to inclined plane, determine the force required

Answer 2

Answer:

The force required is 380N.

Explanation:

The force applied to pull the body along an inclined plane is given as,

$F=mgsin\theta+f$           (1)

Where,

F=net force applied to pull the body

m=mass of the body

g=acceleration due to gravity=10m/s²

f=frictional force acting on the body

From the question we have,

Weight(W)=500N=mg

θ=30°

μ=0.3

The frictional force is given as,

$f=\mu mgcos\theta$           (2)

By substituting the value of μ,  θ, and W in equation (2) we get;

$f=0.3\times 500\times cos30\textdegree$                           ($cos30\textdegree=\frac{\sqrt{3} }{2}$)

$f=150\times \frac{\sqrt{3}}{2}=129.75\approx130N$  (3)

By putting the required values in equation (1) we get;

$F=500\times sin30\textdegree+130=250+130$         ($sin30\textdegree=\frac{1}{2}$ )

$F=380N$

Hence, the force required is 380N.