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Chapter 3: Problem 372

A plate of mass \(\mathrm{M}\) is placed on a horizontal frictionless surface and a body of mass \(m\) is placed on this plate, The coefficient of dynamic friction between this body and the plate is \(\mu\). If a force $2 \mu \mathrm{mg}\(. is applied to the body of mass \)\mathrm{m}$ along the horizontal direction the acceleration of the plate will be (A) \((\mu \mathrm{m} / \mathrm{M}) \mathrm{g}\) (B) \([\mu \mathrm{m} /(\mathrm{M}+\mathrm{m})] \mathrm{g}\) (C) \([(2 \mu \mathrm{m}) / \mathrm{M}] \mathrm{g}\) (D) \([(2 \mu \mathrm{m}) /(\mathrm{M}+\mathrm{m})] \mathrm{g}\)

Short Answer

Expert verified
The acceleration of the plate is (A) \((\mu m / M) g\).

Step by step solution

01

Determining the Maximum Force of Static Friction

Calculate the maximum force of static friction between the plate and the body of mass m. The maximum force of static friction is given by: \( F_{s_{\max}} = \mu m g \)
02

Analyzing Forces on Body m

The horizontal force applied on body m is given as 2μmg. Since this force is greater than the maximum force of static friction, the body will move. The net force acting on body m (Fm) can be calculated as: \( F_m = 2 \mu m g - \mu m g = \mu m g \)
03

Applying Newton's Second Law to Body m

Now that we know the net force acting on body m, we can apply Newton's second law of motion to find its acceleration (am): \( F_{m} = m a_m \) Substitute the values and solve for am: \( a_m = \frac{F_{m}}{m} = \frac{\mu m g}{m} = \mu g \)
04

Analyzing Forces on Plate M

The only horizontal force acting on plate M is the static friction force between the plate and the body of mass m. Since body m is moving to the right, the frictional force will be acting to the left. Therefore, the net force acting on plate M (FM) can be written as: \( F_{M} = -\mu m g \)
05

Applying Newton's Second Law to Plate M

Now that we know the net force acting on plate M, we can apply Newton's second law of motion to find its acceleration (aM): \( F_{M} = M a_{M} \) Substitute the values and solve for aM: \( a_M = \frac{F_{M}}{M} = \frac{-\mu m g}{M} \) The negative sign indicates that the plate accelerates to the left while the body m accelerates to the right. Thus, the acceleration of the plate is (A) \((\mu m / M) g\).

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