The diagram below shows a large cube of
mass 25 kg being accelerated across a frictionless
level floor by a horizontal force, F.
A small cube of mass 4.0 kg is in contact
with the front surface of the cube. The coefficient
of static friction between the cubes
What is the minimum value of F such that the small cube will not slide down the large cube's side?
Physics - 444, Monday, November 25, 2013 at 7:38pm
Physics - Ally, Monday, January 2, 2017 at 7:41pm
Start by drawing the fbd of the little box. Friction up, gravity down, normal towards the right. From that we can tell friction is the same as gravity.
Fg=m•g so Fg=4•9.8
Ff=Fg so Ff=39.2
Ff=μ•Fn so 39.2=.71•Fn
Calculate the acceleration:
Then draw the fbd for the big box. Fn going up, left, and right and Fg going down. Calculate the y axis forces and the Fn going to the left.
Fg=m•g so Fg=25•9.8
Fg=245 and Fn does as well
Fn=55.21 because we solved for that on the small box. Bc of Newton's 3rd law we know those are paired forces.
The answer is 400.28