Question

A force of 100 N needs to be applied parallel to a smooth inclined plane just to hold a body on it. The angle of inclination of the inclined plane is ${30}^o$. How much horizontal force need to be applied to do the same?

• A. 50N
• B. 87N
• C. 100N
• D. 115N

Answer 1

The correct option is D

115N

Lets draw the diagram and see what's happening.

Free Body Diagram of block

Body at equilibrium taking components along and perpendicular and equating net force to 0 we get
$\sum fx=0\sum fy=0$
$\sum fx=0$
100 - mg sin ${30}^o$ = 0
$\Rightarrow$ 100 = mg/2
mg = 200 - - - - - - (1)
New scene
A force is being applied horizontally and its holding block stationary (equilibrium) diagram

Free Body Diagram

Again the body is in equilibrium
$\sum fx=0\sum fy=0$
$\sum fy=0$
$\Rightarrow Ncos{30}^o-mg=0$
N cos 30 = mg

$\sum fx=0$
$\Rightarrow F-Nsin{30}^o=0$
$F=Nsin{30}^o$
$\frac{Nsin{30}^{\circ }}{Ncos{30}^{\circ }}=\frac{F}{mg}\Rightarrow \frac{1}{\sqrt{3}}=\frac{F}{mg}\Rightarrow F=\frac{mg}{\sqrt{3}}=115.4N$