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
is 0.71.
What is the minimum value of F such that the small cube will not slide down the large cube's side?
2 answers
4744
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:
Fg=m•g so Fg=4•9.8
Fg=39.2
Ff=Fg so Ff=39.2
Fn:
Ff=μ•Fn so 39.2=.71•Fn
Fn=55.21126761
Calculate the acceleration:
ΣF=M•A
ΣFx=Fn=m•a
ΣFx=55.21126761=4•a
55.21126761/4=a
a=13.0828169
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:
Fg=m•g so Fg=25•9.8
Fg=245 and Fn does as well
Fn:
Fn=55.21 because we solved for that on the small box. Bc of Newton's 3rd law we know those are paired forces.
ΣF=m•a
ΣFx=Fn-55.12=25•13.08
Fn-55.12= 345.0704225
Fn=400.28
The answer is 400.28
Fg:
Fg=m•g so Fg=4•9.8
Fg=39.2
Ff=Fg so Ff=39.2
Fn:
Ff=μ•Fn so 39.2=.71•Fn
Fn=55.21126761
Calculate the acceleration:
ΣF=M•A
ΣFx=Fn=m•a
ΣFx=55.21126761=4•a
55.21126761/4=a
a=13.0828169
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:
Fg=m•g so Fg=25•9.8
Fg=245 and Fn does as well
Fn:
Fn=55.21 because we solved for that on the small box. Bc of Newton's 3rd law we know those are paired forces.
ΣF=m•a
ΣFx=Fn-55.12=25•13.08
Fn-55.12= 345.0704225
Fn=400.28
The answer is 400.28
6 answers