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# In the figure, a block of weight $60\,N$ is placed on a rough surface. The coefficient of friction between the block and the surface if $0.5$. What should be the weight $W$ such that the block does not slip on the surface.

Last updated date: 05th Mar 2024
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Hint: Any block placed on a rough surface (having some coefficient of friction) moves when the force applied on it is greater than limiting static friction (or limiting friction) whose value is equals to the normal force times the coefficient of friction.

Formula used:
The mathematical expression for static friction is given by,
$f\, = \,\mu N$
where $f$ is the frictional force and $N$ is the normal force.

Let’s analyse the blocks first

Block C: Forces acting on Block C are:
-Normal reaction force from surface on which it is kept (N)
-Tension force (${T_1}$)
-Its weight (60N)
-Static frictional force (f)

As block C is in equilibrium
$\therefore \,\,\sum {{F_x}\, = \,0\,}$
$\Rightarrow \,f - \,{T_{1\,}}\, = \,0$
Hence $f = \,\,{T_{1\,}}$ ……………..…………………$(i)$
also $\sum {{F_y}\, = \,0\,}$
$\Rightarrow \,\,N\, - \,\,60\, = \,0$ or $N\, = \,\,60$ …………..$(ii)$
The frictional force on block C:
$f\, = \,\mu N\, = \,0.5 \times \,60\, = \,30N$………$(iii)$ [using $(ii)$]
Using equation $(i)$ and $(iii)$ we get ${T_1}\,\, = \,\,30N$………….$(iv)$

Block B: Forces acting on Block B are:
Tension force (${T_3}$)

Its weight (W). As block B is in equilibrium
$\therefore \,\sum {{F_y}\, = \,0\,}$ $\Rightarrow \,\,{T_{3\,}}\, - \,\,W\, = \,0$
Hence, ${T_{3\,}}\, = \,\,W$ ………………………………..$(v)$
As the junction P is in equilibrium so at point P
$\sum {{F_x}\, = \,0\,}$
$\therefore \,\,{T_1}\, = \,{T_2} \times \dfrac{1}{{\sqrt 2 }}$……………..$(vi)$
$\sum {{F_y}\, = \,0\,}$
$\therefore \,{T_3}\, = \,\,{T_2}\, \times \dfrac{1}{{\sqrt 2 }}$…………….$(vii)$

Now using equation $(vi)$and$(iv)$we get
${T_1}\, = \,{T_2} \times \dfrac{1}{{\sqrt 2 }}$
$\Rightarrow \,{T_2} = \,\sqrt 2 {T_1}\,$
$\therefore \,\,{T_2}\,\, = \,30\sqrt 2 \,N$……………$(viii)$
Finally, using equation $(v)$,$(vii)$ and $(viii)$
${T_3}\, = \,\,30\sqrt 2 \, \times \dfrac{1}{{\sqrt 2 }}\, = \,30N$ and ${T_{3\,}}\, = \,\,W$

Hence $W\,\, = \,\,30\,Newtons$.

Note: If a block is at rest and keeps on resting on applying force then applied force is equal to the frictional force on the block as net force on the block = 0 in the condition of equilibrium. A body remains at rest if no external force is applied to it.