# Acceleration with two wooden crates

• trolling
In summary, the two wooden crates have a combined mass of 110 kg, with the top crate having a mass of 24 kg and the bottom crate having a mass of 86 kg. There is no friction between the crates and the floor, but there is a coefficient of static friction of 0.8 and a coefficient of kinetic friction of 0.63 between the two crates. A massless rope is attached to the bottom crate, pulling it to the right with a tension of 1122 N. The acceleration of the upper crate is equal to the tension (1122 N) minus the force of kinetic friction (148.176 N), divided by the mass of the top crate (24 kg), resulting in an acceleration of
trolling
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

trolling said:
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

Andrew Mason said:
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.

azizlwl said:
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!

## 1. What is acceleration with two wooden crates?

Acceleration with two wooden crates refers to the process of calculating the change in velocity of two wooden crates over a certain period of time. This can be done by measuring the initial and final velocities of the crates and using the formula a = (vf-vi)/t, where a is the acceleration, vf is the final velocity, vi is the initial velocity, and t is the time interval.

## 2. How do I calculate the acceleration of two wooden crates?

To calculate the acceleration of two wooden crates, you will need to measure the initial and final velocities of the crates and the time interval over which the change in velocity occurred. Then, plug these values into the formula a = (vf-vi)/t and solve for a. The resulting value will be the acceleration of the crates.

## 3. What factors can affect the acceleration of two wooden crates?

The acceleration of two wooden crates can be affected by several factors, including the mass of the crates, the force applied to the crates, and any external forces acting on the crates (such as friction or air resistance). The surface on which the crates are moving can also affect their acceleration.

## 4. How can I use acceleration with two wooden crates in real-life situations?

Acceleration with two wooden crates can be used in real-life situations to understand the motion of objects and to predict how they will move over time. This knowledge can be applied in fields such as engineering, physics, and transportation, where the movement of objects is important.

## 5. Can the acceleration of two wooden crates ever be negative?

Yes, the acceleration of two wooden crates can be negative if their final velocity is less than their initial velocity. This means that the crates are slowing down or decelerating over time. Negative acceleration can also occur if there are external forces acting on the crates that oppose their motion, such as friction or air resistance.

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