Question

A weight of $500\,$N is held on a smooth plane inclined at $30^\circ$ to the horizontal by a force $P$ acting at $30^\circ$ to the inclined plane as shown. Then the value of force $P$ is: 

• A. $(500/\sqrt{3})\,\text{N}$
• B. $(500\sqrt{3})\,\text{N}$
• C. $500(\sqrt{3}/2)\,\text{N}$
• D. $250\,\text{N}$

Hint:

On smooth planes, set $\sum F_{\parallel}=0$ using only the components of applied forces and weight; the normal reaction has no tangential component.

Answer 1

The Correct Option is A

 

Step 1: Resolve along and normal to the plane.
Let the plane angle be $\theta=30^\circ$, the force $P$ is inclined $\alpha=30^\circ$ above the plane.
Weight components: along the plane $W\sin\theta=500\sin30^\circ=250$ (down-slope); normal $W\cos\theta=500\cos30^\circ=250\sqrt{3}$.
Force $P$ components: along the plane $P\cos\alpha$ (up-slope); normal $P\sin\alpha$ (away from plane).

Step 2: Equilibrium along plane (no friction).
\[ P\cos 30^\circ = W\sin 30^\circ \Rightarrow P\left(\frac{\sqrt{3}}{2}\right)=250 \Rightarrow P=\frac{250}{\sqrt{3}/2}=\frac{500}{\sqrt{3}}\,\text{N}. \]

Step 3: Check normal reaction is positive.
\[ R = W\cos 30^\circ - P\sin 30^\circ = 250\sqrt{3} - \frac{500}{\sqrt{3}}\cdot\frac{1}{2} = \frac{500}{\sqrt{3}} > 0 (\text{OK}). \]

Step 4: Conclusion.
$P=\dfrac{500}{\sqrt{3}}\,\text{N}$.