How come no one seems to know how to work this problem?

Question

Please help with this physics problem.

A 4.00 kg block is placed on top of a 12.0 kg block that rests on a frictionless table. The coefficient of static friction between the two blocks is 0.700. What is the maximum horizontal force that can be applied before the 4.00 kg block begins to slip relative to the 12.0 kg block, if the force is applied to each of the following?

(a) the more massive block

= 109 (.7 * 16*9.8)

(b) the less massive block

How do I do part b?

The answer for part b is not, 53.85, 82.32, 27.44, or 109.

Answer 1

Let mu 0.700, M1 = 4 kg, M2 = 12 kg, g = 9.8 meters/sec^2

Let f be the magnitude of the lateral force of friction applied by each block upon the other one.

If upper block is sliding on the lower block, then f = mu*g*M1
This means if not sliding, f
Also, not sliding means lateral accelerations a1 and a2 are equal for the two blocks.

Part (a), where Lateral force F is applied to M2:
a2 = (F-f)/M2, and a1 = f/M1, so (F-f)/M2 = f/M1
M1*F - M1*f = M2*f, (M2+M1)*f = M1*F
f = F * M1/(M2+M1), and remember that f < mu*g*M1,
so the maximum F is found from F*M1/(M2+M1) < mu*g*M1
and thus F < mu*g*(M2+M1)
F < 0.700*9.8*16 = 109.76 kg*meter/sec^2


Part (b), where lateral force F is applied to M1:
a1 = (F-f)/M1, and a2 = f/M2, so (F-f)/M1 = f/M2
M2*F - M2*f =M1*f, (M1+M2)*f = M2*F
f = F * M2/(M1+M2), and remember that f < mu*g*M1
so the maximum F is found from F*M2/(M1+M2) < mu*g*M1
and thus F < mu*g*M1*(M1+M2)/M2
F < 0.700*9.8*4*(4+12)/12 = 0.700*9.8*16/3
F < 36.59 kg*meter/sec^2

Answer 2

There is a maximum allowable force between the two blocks... when you push hard enough on the upper block to slip between the blocks it means that the force to accelerate the 12 kg block is just greater than the static friction force.

So:

4kg * 9.8 N/kg * .7 = 27.44 Newtons.

The force to break the static is 27.44....

If you push on the 4kg block any harder than that, it will not be able to transfer the force to the 12 kg block.

why do you need to go further?

Answer 3

"IF" I understand the question correctly...

4 kg (weight of the top block)

.70 coefficient of friction

4 * .70 = 2.8 kg

2.8 kg of force greater or lesser than that which is applied to the bottom brick will cause them to slide in relation to each other.

How you manage to introduce that stress where the two blocks meet wasn't specified nor was it stated that a description of such a method was necessary.

Answer = 2.8 kg

Trick question. :-)

Answer 4

The hang-up is the frictionless surface that the 12 kg block is setting on. Any force applied to either block will cause both to move as a unit.

Answer 5

Yup, At the moment i feel as big as an Elephant, I'm full of flu and i have a very bad Cough. Does that count?