Question

The drawing shows a large cube (mass = 25 kg) being accelerated across a horizontal frictionless surface by a horizontal force . A small cube (mass = 4.0 kg) is in contact with the front surface of the large cube and will slide downward unless is sufficiently large. The coefficient of static friction between the cubes is 0.71. What is the smallest magnitude that can have in order to keep the small cube from sliding downward?

Answer 1

the answer is 400 N

Answer 2

first we find the normal force on the small cube:
F=ma (where f=force,a=acceleration,m=mass)
a=f/m
acceleration on small box is same as that on the large box. let force to find be P.
then:
a=p/(25kg+4kg)
a=p/(29kg)m/s^2
force acting on small box:
f=ma
f=4*(p/29)N(normal force)
friction force= Us*(normal force) Us is coeff. of static friction.

friction= 0.71*(4*p/29)
weight= mg= 4*9.8
for the object(small box) not to slide down the friction force b/w the two objects have to be exactly the same as the weight of the object.
0.71*(4*p/29)=4*9.8
solving..
p= 400.29N

Answer 3

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

Answer 4

will there be Ff as reaction force going vertically down on the big box?