# Acceleration with two wooden crates

1. Homework Statement

Two wooden crates rest on top of one another. The smaller top crate has a mass of m1 = 24.0 kg and the larger bottom crate has a mass of m2 = 86.0 kg. There is NO friction between the crate and the floor, but the coefficient of static friction between the two crates is μs = 0.8 and the coefficient of kinetic friction between the two crates is μk = 0.63. A massless rope is attached to the lower crate to pull it horizontally to the right (which should be considered the positive direction for this problem).

The tension is increased in the rope to 1122.0 N causing the boxes to accelerate faster and the top box to begin sliding. What is the acceleration of the upper crate?

2. Homework Equations

Fg = Fn = m*g
Fk = μk * Fn
Fnet,x = T-Fk

3. The Attempt at a Solution

Fg = 24*9.8 = 235.2N
Fn = 235.2N
Fk = .63 * Fn = 148.176N
T = 1122N
Fnet,x = 1122-148.176 = 973.824N
acceleration in x-direction = 973.824/24 = 40.576 m/s^2 <- seems to be incorrect

Andrew Mason
Homework Helper
1. Homework Statement

Two wooden crates rest on top of one another. The smaller top crate has a mass of m1 = 24.0 kg and the larger bottom crate has a mass of m2 = 86.0 kg. There is NO friction between the crate and the floor, but the coefficient of static friction between the two crates is μs = 0.8 and the coefficient of kinetic friction between the two crates is μk = 0.63. A massless rope is attached to the lower crate to pull it horizontally to the right (which should be considered the positive direction for this problem).

The tension is increased in the rope to 1122.0 N causing the boxes to accelerate faster and the top box to begin sliding. What is the acceleration of the upper crate?

2. Homework Equations

Fg = Fn = m*g
Fk = μk * Fn
Fnet,x = T-Fk

3. The Attempt at a Solution

Fg = 24*9.8 = 235.2N
Fn = 235.2N
Fk = .63 * Fn = 148.176N
T = 1122N
Fnet,x = 1122-148.176 = 973.824N
acceleration in x-direction = 973.824/24 = 40.576 m/s^2 <- seems to be incorrect
You are asked for the acceleration of the upper crate. This is the acceleration of the lower (86 kg) crate. Also, express your answer in proper significant figures.

AM

You are asked for the acceleration of the upper crate. This is the acceleration of the lower (86 kg) crate. Also, express your answer in proper significant figures.

AM

So how do you suggest that you find the upper crate's acceleration?

To me the only force applied to the top block is just mgμk since it is sliding.
Thus maximum acceleration equal to gμk.

For no sliding, maximum acceleration is gμs.

To me the only force applied to the top block is just mgμk since it is sliding.
Thus maximum acceleration equal to gμk.

For no sliding, maximum acceleration is gμs.

I just typed it in, & it's correct. Thanks for your help!