Question

Find the least pulling force acting at an angle of ${30}^{\circ }$ with the horizontal, which can just slide a body of mass $10\text{kg}$ along a rough horizontal surface $[{\mu }_s={\mu }_k=\frac{1}{\sqrt{3}},g=10m/s^2]$

• A. $25\text{N}$
• B. $75\text{N}$
• C. $50\text{N}$
• D. $100\text{N}$

Answer 1

The correct option is C $50\text{N}$

From the FBD of body

Along horizontal direction,
$f=P\cos {30}^{\circ }--\left(1\right)$
Along vertical direction,
$R+P\sin {30}^{\circ }=10g$

When body is just sliding,
$f={\mu }_{k}R={\mu }_k(10g-P\sin {30}^{\circ })--\left(2\right)$

From (1) and (2),
$P\cos {30}^{\circ }={\mu }_k(100-P\sin {30}^{\circ })$
$P(\cos {30}^{\circ }+{\mu }_k\sin {30}^{\circ })={\mu }_k\left(100\right)$
$P=\frac{{\mu }_k\times 100}{\cos {30}^{\circ }+{\mu }_k\sin {30}^{\circ }}$
$=\frac{\frac{100}{\sqrt{3}}}{\frac{\sqrt{3}}{2}+\frac{1}{2\sqrt{3}}}$
$=\frac{200}{4}=50\text{N}$