- #1

Anne Armstrong

- 11

- 1

## Homework Statement

Two wooden crates rest on top of one another. The smaller top crate has a mass of m

_{1}= 20 kg and the larger bottom crate has a mass of m

_{2}= 80 kg. There is NO friction between the crate and the floor, but the coefficient of static friction between the two crates is μs = 0.79 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). Tension is initially 232 N. If the tension is increased in the rope to 1085 N causing the boxes to accelerate faster and the top box to begin sliding. What is the acceleration of the upper crate?

## Homework Equations

F

_{fr}=μmg

F=ma

## The Attempt at a Solution

Based on previous questions, I found that the maximum acceleration at which the crates can move without the top crate sliding is 7.74 m/s

^{2}and the maximum tension at which the crates can be pulled without the top crate sliding is 46.4 N.

I attempted to use the same approach to solve this problem:

F=m

_{1,2}*a --> 1085 N = (80 kg + 20 kg)*a --> a=10.85 m/s

^{2}

F=m

_{1}*a --> F=(20 kg)*(10.85 m/s

^{2}) --> F=217 N =the net force on the top crate

F

_{net}=F

_{T}-F

_{fr}or F

_{T}-F

_{net}=F

_{fr}

so 1085 N - 217 N = F

_{fr}--> F

_{fr}=868 N

F

_{fr}=m

_{1}*a --> 868 N = (20 kg)*a --> a=43.4 m/s

^{2}

I have a feeling there is an error in the first step (finding the acceleration to be 10.85 m/s

^{2}, but I'm not sure. I also realized after I finished that I didn't include any coefficients of friction, which I don't think is correct... I hope my process was clear, I'd appreciate any help! Thanks