Alarge cube has a mass of 25kg. it is being acclerated
acrossa honzontial frictionless surface by a horizontal force. a
smallcube with a mass of 4kg is in contact with the front of
thelareg cube and will slide downward unless the horizontal forceis
sufficiently large. the coefficient of static friction betweenthe
two cubes is 0.71. what is the smallest magnitude that
thehorizontal force can have in order to keep the small cube
fromsliding downward?

p= 400.29N the horizontal force

Explanation:

Given data

mass=25 kg

small cube mass mass=4 kg

Us (The Coefficient of static b/w two cubes) = 0.71

to find

The horizontal force to keep the small cube  from sliding downward

Solution

F=mafrom Newton Second law

Where F=force

a=acceleration

m=mass

we can write equation in form of acceleration  

a=F/m

The acceleration on small box is same as that on the large box.

Let P be force to find.

then:

a=p/(25kg+4kg)

a=p/(29kg)m/s²

The force acting on small box:

F=ma

f=4*(p/29)Nnormal force

friction force= Us*(normal force)where Us is coefficient of static friction.

friction force= 0.71*(4*p/29)

Now to find weight

weight= mg

weight= 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 for p(force)

p= 400.29N the horizontal force

force is mass*acceleration.

we have a very small mass: 11.4*0.001 kg

if we want 0.035n of force, we need an acceleration:

0.035 = 11.4*0.001*a

a = 0.035/(11.4*0.001) m/s^2