- #1
Anne Armstrong
- 11
- 1
Homework Statement
Two wooden crates rest on top of one another. The smaller top crate has a mass of m1 = 20 kg and the larger bottom crate has a mass of m2 = 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
Ffr=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/s2 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=m1,2*a --> 1085 N = (80 kg + 20 kg)*a --> a=10.85 m/s2
F=m1*a --> F=(20 kg)*(10.85 m/s2) --> F=217 N =the net force on the top crate
Fnet=FT-Ffr or FT-Fnet=Ffr
so 1085 N - 217 N = Ffr --> Ffr=868 N
Ffr=m1*a --> 868 N = (20 kg)*a --> a=43.4 m/s2
I have a feeling there is an error in the first step (finding the acceleration to be 10.85 m/s2, 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