Question

A $3\text{kg}$ block is pulled by a force which is inclined at ${37}^{\circ }$ to the horizontal table. The friction coefficient between the table and block is 1/3. For what minimum value of this force, will the block start sliding?

• A. $5\text{N}$
• B. $10\text{N}$
• C. $20\text{N}$
• D. $25\text{N}$

Answer 1

The correct option is B $10\text{N}$

Free body diagram of the given block can be given as

If we have calculate the minimum force to just slide
$F_{min}\cos {37}^{\circ }=f_s$
where
$f_s$ is the static friction
$\Rightarrow F_{min}\cos {37}^{\circ }=\mu N$ ....(i)
Now balancing the vertical force we get
$N+F_{min}\sin {37}^{\circ }=mg\Rightarrow N+\frac{3}{5}{F}_{min}=30\Rightarrow N=30-\frac{3}{5}{F}_{min}$
Now putting this in eq(i)
$\frac{4}{5}{F}_{min}=\frac{1}{3}(30-\frac{3}{5}{F}_{min}$
On solving this we get
$F_{min}=10\text{N}$