A 15 kg box is resting on a ramp inclined at 40 degrees. If the coefficient of static friction between the box and the ramp is 0.5, what is the maximum angle of inclination at which the box can remain at rest without sliding down?

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Seekh · Accepted Answer

The box will stay at rest as long as the downhill component of its weight, $mg\sin\theta$, does not exceed the maximum static friction $f_{\text{max}}=\mu_s mg\cos\theta$. Setting $mg\sin\theta \le \mu_s mg\cos\theta$ and cancelling $mg$ gives $\tan\theta \le \mu_s$. With a static‑friction coefficient of 0. 5, the largest angle that satisfies this is \(\theta_{\max}=\arctan(0. 5)\approx 26.

A 15 kg box is resting on a ramp inclined at 40 degrees. If the coefficient of static friction between the box and the ramp is 0.5, what is the maximum angle of inclination at which the box can remain at rest without sliding down?

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