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:
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On smooth planes, set $\sum F_{\parallel}=0$ using only the components of applied forces and weight; the normal reaction has no tangential component.
Step 1: Resolve forces relative to the plane. Given plane angle $\theta=30^\circ$ and force $P$ inclined $\alpha=30^\circ$ above the plane. Weight components: along the plane $W\sin\theta=500\sin30^\circ=250$ (downwards); perpendicular to the plane $W\cos\theta=500\cos30^\circ=250\sqrt{3}$. Force $P$ components: along the plane $P\cos\alpha$ (upwards); perpendicular to the plane $P\sin\alpha$ (away from the plane).
Step 2: Analyze equilibrium along the plane (assuming 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: Verify the normal reaction is positive. \[ R = W\cos 30^\circ - P\sin 30^\circ = 250\sqrt{3} - \frac{500}{\sqrt{3}}\cdot\frac{1}{2} = \frac{500}{\sqrt{3}} > 0 (\text{Consistent}). \]
Step 4: State the final result. $P=\dfrac{500}{\sqrt{3}}\,\text{N}$.
