Answer to Question #132548 in Mechanics | Relativity for Shruti

Question

determine the horizontal force P required to start moving the 500N block up the inclined surface.assume static friction is 0.3 and angle is 30

Answer 1

On this picture $\mathbf{mg}$ denotes the gravity force, $\mathbf{F_{fric}} = \mu\mathbf{N}$ - static frictional force and $\mathbf{N}$ - normal force. According to the problem, $mg = 500 N$, $\mu = 0.3$ and $\theta = 30\degree$.

According to the Newton's second law, the projections on the y axis will be:

$N - mg\cos\theta - P\sin\theta = 0\\ N =mg\cos\theta + P\sin\theta$

The projections on the x axis will be:

$F_{fric} + mg\sin\theta - P\cos\theta = 0\\ \mu N + mg\sin\theta - P\cos\theta = 0\\$

Substituting $N$ from the first expression, get:

$\mu (mg\cos\theta + P\sin\theta) + mg\sin\theta - P\cos\theta = 0$

Expressing $P$, obtain:

$P = mg\dfrac{\mu + \tan\theta}{1 + \mu\tan\theta}\\ \space\\ P = 500\cdot \dfrac{0.3 + \tan30\degree}{1 + 0.3\cdot \tan30\degree} \approx 373.9 N$

Answer. 373.9 N.