In figure, the mass m2 = 10 kg slides on a frictionless shelf. The coefficients of static and kinetic friction between m2 and m1 = 5 kg are µs = 0.6 and µk = 0.4.

(a) What is the maximum acceleration of m1?
(b) What is the maximum value of m3 if m1 moves with m2 without slipping?
(c) If m3 = 30 kg, find the acceleration of each body and the tension in the string.

M1
M2-----------
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m3

To find the maximum acceleration of m1, we need to consider the forces acting on it. The only force acting on m1 is the force of friction between m1 and m2. Since m1 is on the verge of slipping, the friction force is at its maximum, which is equal to the product of the coefficient of static friction (µs) and the normal force (N). The normal force is equal to the weight of m1, which is m1 multiplied by the acceleration due to gravity (g).

(a) Maximum acceleration of m1:
Friction force = µs * N = µs * (m1 * g)

Next, we need to consider the net force acting on m1. It is given by the equation:
Net force = m1 * acceleration

Since there is only one force acting on m1, which is the friction force, the net force is equal to the friction force. So, we have:
µs * (m1 * g) = m1 * acceleration

Simplifying the equation, we get:
acceleration = µs * g

Substituting the given values, we find:
acceleration = 0.6 * 9.8 m/s^2
acceleration ≈ 5.88 m/s^2

Therefore, the maximum acceleration of m1 is approximately 5.88 m/s^2.

(b) To find the maximum value of m3 for m1 to move with m2 without slipping, we need to consider the forces acting on m3. The maximum force of static friction between m2 and m3 is equal to the product of the coefficient of static friction (µs) and the normal force (N). The normal force is equal to the weight of m3, which is m3 multiplied by the acceleration due to gravity (g).

Maximum force of static friction = µs * N = µs * (m3 * g)

To prevent m1 from slipping, the maximum force of static friction between m2 and m3 should be greater than or equal to the force of kinetic friction between m1 and m2. The force of kinetic friction is equal to the product of the coefficient of kinetic friction (µk) and the normal force (N).

Force of kinetic friction = µk * N = µk * (m1 * g)

Setting up the inequality:
µs * (m3 * g) ≥ µk * (m1 * g)

Simplifying the inequality, we find:
m3 ≥ (µk / µs) * m1

Substituting the given values, we get:
m3 ≥ (0.4 / 0.6) * 5 kg
m3 ≥ 2/3 * 5 kg
m3 ≥ 10/3 kg
m3 ≥ 3.33 kg

Therefore, the maximum value of m3 for m1 to move with m2 without slipping is approximately 3.33 kg.

(c) To find the acceleration of each body when m3 = 30 kg and the tension in the string, we need to consider the forces acting on each body.

For m1:
The force of friction between m1 and m2 is given by the equation:
Force of friction = µk * N = µk * (m1 * g)

The net force acting on m1 is given by the equation:
Net force = m1 * acceleration

Since there is only one force acting on m1, which is the force of friction, the net force is equal to the force of friction. So, we have:
µk * (m1 * g) = m1 * acceleration

Simplifying the equation, we get:
acceleration = µk * g

Substituting the given values, we find:
acceleration = 0.4 * 9.8 m/s^2
acceleration ≈ 3.92 m/s^2

For m2 and m3:
Since they are connected by a string, they will have the same acceleration.

Therefore, the acceleration of m2 and m3 is also approximately 3.92 m/s^2.

To find the tension in the string, we can consider the forces acting on m2 and m3. The tension in the string is equal to the force exerted by m2 on m3.

Using Newton's second law, we can write the equation:
Force = mass * acceleration

For m2:
Force = m2 * acceleration = 10 kg * 3.92 m/s^2

For m3:
Force = m3 * acceleration = 30 kg * 3.92 m/s^2

Therefore, the tension in the string is equal to the force exerted by m2 on m3, which is approximately 39.2 N.