How Does Friction Affect the Acceleration of Boxes on a Ramp?

[abel2]

Homework Statement

Two boxes are connected by a weightless cord that runs through a frictionless pulley. Box B is hanging directly down from the pulley, while Box A is located on a ramp with a coefficient of kinetic friction at 0.15. The two boxes have the same mass of 2.7kg. What is the acceleration of Box B if the ramp has an angle of 34 degrees?

Homework Equations

F(friction) = F(normal) * coefficient of kinetic friction for Box A
F(normal) = m*g for Box B

The Attempt at a Solution

After calculating the normal force for Box A at (2.7 kg) * (9.8 m/s/s) * sin 124 = 21.94 N, then the F(friction) = 21.94 * 0.15 = 3.29 N
The normal force for Box B is (2.7 kg ) * g = 26.46 N

So Box B will pull down on Box A with a force of 26.46 - (21.94 + 3.29) = 1.23 N or 1.23 / 5.4kg = 0.23 m/s/s

I think this is correct since it makes sense numerically. The acceleration wouldn't be that large so 0.23 fits. Sometimes problems like this cause me to over think them a lot and I end up doubting my answers. Any verification would be appreciated.

 

[tiny-tim]

welcome to pf!

hi abel2! welcome to pf!

After calculating the normal force for Box A at (2.7 kg) * (9.8 m/s/s) * sin 124 = 21.94 N, then the F(friction) = 21.94 * 0.15 = 3.29 N
The normal force for Box B is (2.7 kg ) * g = 26.46 N

So Box B will pull down on Box A with a force of 26.46 - (21.94 + 3.29) = 1.23 N …

no, you've used the normal force (21.94), you need the slope-component of the weight of Box A