Question

A block of mass M= 10 kg is placed on an inclined plane, inclined at angle $\theta ={37}^0$ with the horizontal. The coefficient of friction between the block and inclined is $\mu =0.5$.
a. Calculate the acceleration of the block when it is released.
b. Now a force F= 75 N is applied on block as shown. Find out the acceleration of the block. If the block is initially at rest.
c. In case (b), how much force should be added to 75 N of force so that block the starts to move up the incline.
d. What is the minimum force by which 75N force should be replaced with so that block does not move?

Answer 1

Here angle of repose $\alpha =ta{n}^{-1}\left({\mu }_s\right)$

$\alpha =ta{n}^{-1}\left(\frac{1}{2}\right)={30}^0$

The angle of inclination is greater than the angle of respose. The friction force on the block will act in upward direction.

For the acceleration of block,

$Mgsin\theta -\mu N=Ma$

$\Rightarrow a=gsin\theta -\mu gcos\theta$

$=10(sin{37}^0-0.5cos{37}^0)$

b. If external force F= 75N is applied on the block.

Let us find net driving force acting on block. Parallel to inclined two external forces are acting one in upward direction F and other is the component of weight in the direction downward the plane, $Mgsin\theta$.

Net driving force $f_{driving}=F-Mgsin\theta$

$\Rightarrow F_{driving}=75-10\times 10\times sin{37}^0=75-60=15N$

Maximum resisting force that oppose relative motion is maximum friction force (or $f_{lim}$)

$f_{lim}={\mu }_{s}Mgcos\theta$

$=0.5\times 10\times 10\times cos{37}^0=40N$

Here $f_{lim}<f_{resisting}$. Hence, the block will not move and friction will be static will act in the direction opposite to driving force, i.e., in downward direction.

c. To move block, the least value of driving force should be 40 N. But in above case, driving force is 15N (up). Hence, if we add $△F=25N$ in upward direction, the block will overcome maximum resistence force (or friction) and starts moving up.

$\therefore 60+40=75+△F\Rightarrow △F=25N$

d. As resisting force which is maximum friction force is 40N and the component of weight parallel to incline is 60 N and acting downward. If we remove F, then the driving forces will be the only component of the weight in the direction downward the incline plane. In this case, friction will act in upward direction. Hence, the required value of F to make block in equilibrium,

F + 40 = 60

or F = 20N