The coefficient of static friction between a | vedclass

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(A) The Free Body Diagram $(FBD)$ of the block is shown in the diagram. Since the block is at rest,the forces must be balanced. Vertical direction: Th

Vedclass · Accepted Answer

$25$

Vedclass · Answer

(A) The Free Body Diagram $(FBD)$ of the block is shown in the diagram. Since the block is at rest,the forces must be balanced. Vertical direction: The frictional force $f_r$ balances the weight $mg$. $f_r - mg = 0 \Rightarrow f_r = mg$ $........(1)$ Horizontal direction: The applied force $F$ is balanced by the normal reaction $N$ from the wall. $F - N = 0 \Rightarrow N = F$ $..........(2)$ We know that the frictional force $f_r \leq \mu N$. In the limiting case,to keep the block from sliding down: $f_r = \mu N$ Substituting $f_r = mg$ and $N = F$ into the equation: $mg = \mu F$ $F = \frac{mg}{\mu}$ Given $m = 0.5\, kg$,$g = 10\, m/s^2$,and $\mu = 0.2$: $F = \frac{0.5 	imes 10}{0.2} = \frac{5}{0.2} = 25\, N$.

The coefficient of static friction between a wooden block of mass $0.5\, kg$ and a vertical rough wall is $0.2$. The magnitude of the horizontal force that should be applied on the block to keep it adhered to the wall will be $N$ $\left[ g = 10\, m/s^2 \right]$.

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