Question

A pulling force of $100\text{N}$ is acting at an angle of ${30}^0$with the horizontal on a body of mass $50\text{kg}$ placed on a rough horizontal surface. Will the same force cause the body to move if it is a pushing force for the body? If not, what should be the value of the force (acting at the same angle) to cause the body to just move? Take $g=10{\text{m/s}}^2$.

• A. $77\text{N}$
• B. $124.84\text{N}$
• C. $88\text{N}$
• D. $119\text{N}$

Answer 1

The correct option is B $124.84\text{N}$


From the FBD, when the force is pulling the body.
Applying equations for the equilibrium of the body in horizontal and vertical direction:

For just moving condition, $f_{max}={\mu }_{s}N$ will act on the body.

$N=mg-F\sin {30}^0.....\left(i\right)F\cos {30}^0={\mu }_{s}N.....\left(ii\right)$

Substituting the values, we get:
$F\cos {30}^{\circ }={\mu }_s(mg-F\sin {30}^{\circ })$
$F=\frac{{\mu }_{s}mg}{\cos {30}^{\circ }+{\mu }_s\sin {30}^{\circ }}$
$\Rightarrow 100=\frac{{\mu }_s\times 50\times 10}{\frac{\sqrt{3}}{2}+{\mu }_s\times \frac{1}{2}}$
$\Rightarrow {\mu }_s=0.192$

FBD in case of pushing force (with same force at same angle with horizontal):


Following same procedure for the pushing case:

$N=mg+F\sin {30}^0.....\left(i\right)F\cos {30}^0={\mu }_{s}N.....\left(ii\right)$

Substituting (i) in (ii), we get:
$F=\frac{{\mu }_{s}mg}{\cos {30}^{\circ }-{\mu }_s\sin {30}^{\circ }}$
$F=\frac{500\times 0.192}{0.769}=124.84\text{N}$