Two crates, of mass m1 = 40 kg and m2 = 125 kg, are in contact and at rest ?

Question

on a horizontal surface. A 620 N force is exerted on the 40 kg crate. The coefficient of kinetic friction is 0.15. Calculate the acceleration of the system.m/s2 (to the right)

Calculate the force that each crate exerts on the other

Repeat with the crates reversed

(acceleration=m/s2)

crate force=N

Answer 1

Applied force Fapp = 620 N

Consider the two forces together as system.

Mass m = m1+ m2 = 40 + 125 = 165 kg

Forces on the system are

1. Weight mg downward

2. Normal force N by the table upward

3. Friction uN by the table opposite to the direction in which crates move (u = coefficient of friction)

4. Applied force Fapp in the direction of motion.

Normal force balances gravity.

Therefore, N = mg

Total horizontal force F = Fapp - uN

F = Fapp - uMg

F = 620 - 0.15 * 165 * 9.8 = 620 - 242.55 = 422.45 N

Acceleration a = F/m = 422.55/165 = 2.56 m/s^2

Consider m1 as system. In horizontal direction, there are three forces acting on it

1. Applied force 620 N

2. Friction um1g = 0.15 * 40 * 9.8 = 58.8 N opposite to applied force

3. Force F2 by m2 opposite to applied force

Total force = 620 - 58.8 - F2 = 561.2 - F2

Force = mass * acceleration

561.2 - F2 = 40 * 2.56 = 102.4

F2 = 561.2 - 102.4 = 458.8 N

Ans:

Acceleration = 2.56 m/s^2

Crates exert 458.8 N on each other.

You can solve for the crates reversed in the same way. Use m1 where I have used m2 and use m2 where I have used m1.

Answer 2

OK let T be the force acting between the masses. By Newton's law, this force has an equal and opposite reaction on each of them.

For each of them, the normal reaction is equal to its weight, and the friction equal to μ*mg

Forces on the 40 kg mass:

620 - T - 40*0.15*9.81 = 40*a ...(1)

where a = common acceleration of both masses

Forces on the 125 kg mass:

T - 125*0.15*9.81 = 125*a ...(2)

Now you need to solve (1) and (2) simultaneously to get a and T.

a = 2.288 m/s^2

T = 469.7 N

Reversing the crates gives

Forces on the 125 kg mass:

620 - T - 125*0.15*9.81 = 125*a ...(1)

where a = common acceleration of both masses

Forces on the 40 kg mass:

T - 40*0.15*9.81 = 40*a ...(2)

Now you need to solve (1) and (2) simultaneously to get a and T.

a = 2.288 m/s^2

T = 150.3 N