Question

Two crates, of mass 65 kg and 125 kg, are in contact and at rest on a horizontal surface (fig. 4-57). A 650 N force is exerted on the 65 kg crate. If the coefficient of kinetic friction is 0.18, calculate:

(a) the acceleration of the system, and
(b) the force that each crate exerts on the other.
(c) Repeat with the crates reversed.

Answer 1

(a) The acceleration of the system is
`\( \mathbf{1.23 \, m/s^2} \).`

(b) The force exerted by each crate on the other is
`\( \mathbf{482.5 \, N} \).`

(c) When the crates are reversed, the acceleration remains the same at
`\( \mathbf{1.23 \, m/s^2} \),` and the force exerted by each crate on the other is
`\( \mathbf{482.5 \, N} \).`

Step-by-step explanation:

To calculate the acceleration
`(\(a\))`, we can use Newton's second law:

`\[ \Sigma F = m \cdot a \]`

The net force acting on the system
`(\(\Sigma F\))` is the applied force minus the force of friction. The force of friction
`(\(\Sigma F\))` can be calculated using the coefficient of kinetic friction
`(\(\mu_k\)):`

`\[ f_k = \mu_k \cdot N \]`

where \(N\) is the normal force. The normal force is equal to the weight of the crates
`(\(m \cdot g\))` since the system is on a horizontal surface.

Once we find
`\(f_k\),` we subtract it from the applied force to get the net force. Finally, we use Newton's second law to find the acceleration.

For part (b), the force each crate exerts on the other is equal to the force of friction
`(\(f_k\))` because the system is at rest.

For part (c), reversing the crates doesn't change the physics involved. The normal force, friction force, and net force remain the same, resulting in the same acceleration and force between the crates.