Accelerating blocks

  • #1

Homework Statement



Two wooden crates rest on top of one another. The smaller top crate has a mass of m1 = 22 kg and the larger bottom crate has a mass of m2 = 94 kg. There is NO friction between the crate and the floor, but the coefficient of static friction between the two crates is μs = 0.75 and the coefficient of kinetic friction between the two crates is μk = 0.6. 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 rope is pulled with a tension T = 239 N (which is small enough that the top crate will not slide).

What is the frictional force the lower crate exerts on the upper crate?



Homework Equations



Ffr=μmg
Fnet=ma




The Attempt at a Solution



if F(net)=T, then ma=T, and (94+21)a=239, a=2.08m/s2.

I really am not sure how to proceed from here to find the force of friction exerted by the lower block on the upper block....I would really appreciate some help! Thanks!
 

Answers and Replies

  • #2
PeterO
Homework Helper
2,426
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Homework Statement



Two wooden crates rest on top of one another. The smaller top crate has a mass of m1 = 22 kg and the larger bottom crate has a mass of m2 = 94 kg. There is NO friction between the crate and the floor, but the coefficient of static friction between the two crates is μs = 0.75 and the coefficient of kinetic friction between the two crates is μk = 0.6. 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 rope is pulled with a tension T = 239 N (which is small enough that the top crate will not slide).

What is the frictional force the lower crate exerts on the upper crate?



Homework Equations



Ffr=μmg
Fnet=ma




The Attempt at a Solution



if F(net)=T, then ma=T, and (94+21)a=239, a=2.08m/s2.

I really am not sure how to proceed from here to find the force of friction exerted by the lower block on the upper block....I would really appreciate some help! Thanks!
The expression:
Ffr=μmg

enables you to calculate the maximum possible friction, but friction is only ever as large as it needs to be.

You have accurately calculated the acceleration of the crates, you now need to consider what force is needed to accelerate the top crate, and where that force comes from.
(hint: if it is not gravity, magnetism or electrostatics - it must be a contact force, so consider everything that is touching the top crate)
 

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