# Forces acting on a block which is lying on another block

• Like Tony Stark
Like Tony Stark
Homework Statement
In the picture, the coefficient of static and dynamic friction for all contact surfaces are ##0.4## and ##0.14##, respectively. The mass of ##A## is ##73 kg## and the mass of ##B## is ##22 kg##. Determine the maximum force so that B doesn't slide. Then, suppose that the force applied to the system is greater than the force calculated previously, how much time would it take for the blocks to move ##10 m##?
Relevant Equations
Newton's equations
Well, I'm having trouble with the free body diagrams. For ##A## we have
##y)## weight, normal force, contact force with ##B##, ##F . sin(36.8°)##. And the acceleration is ##0## because we want to calculate the maximun force before moving.

##\Sigma \vec F = m . a_y##
##\vec N_A + \vec F . sin(36.8°) - \vec W_A -\vec Fc= 0##

##x)## ##F . cos(36.8°)##, friction force with the ground and friction force with ##B## and the acceleration is 0.

##F. cos(36.8°) - Fr - Fr_B=0##

##Fr## and ##Fr_B## have the same direction, don't they? Because the friction with ##B## must be pushing ##B## towards, and then the reaction force is in the opposite direction.

For ##B## we have:
##y)## weight and contact force (normal force)

##\vec N_B - \vec W_B =0##

##x)## friction force with ##A##
But is this force pointing to the right? And then, is the string applying a force on ##B##?

And then when I want to answer the second question, do they move equally? I mean, if ##A## traveled ##10 m## so did ##B##

#### Attachments

• 20190927_085748.jpg
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Homework Helper
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normal force, contact force with B
The contact force with B could include a frictional component in general. Better to say normal force from ground, normal force from B.
the acceleration is 0 because we want to calculate the maximun force before moving.
No, it's the maximum force before B slides on A. A will be moving by then.

Like Tony Stark
The contact force with B could include a frictional component in general. Better to say normal force from ground, normal force from B.

No, it's the maximum force before B slides on A. A will be moving by then.

So the free body diagram for A is
##x) F.cos(36,8°)-Fr-Fr_{AB}=m.a_x##
##y) F.sin(36,8°)+N-P-F_{AB}=m.a_y##

And for B is
##x) Fr_{AB}=m.a_x##
##y) F_{AB}-P=m.a_y##

Is this correct?

And for the ##y)## axis of both bodies the acceleration is equal to 0, isn't it? Because they don't move upwards. Is this right?

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Is this correct?
Yes, except what is P in these equations? Not the same in both, I assume.
And for the y) axis of both bodies the acceleration is equal to 0
Yes.

Like Tony Stark
Yes, except what is P in these equations? Not the same in both, I assume.

Yes.
Sorry, ##P## is weight
Thanks!

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Sorry, ##P## is weight
Thanks!
Ok, but different weights for the two blocks, right?

Like Tony Stark
Ok, but different weights for the two blocks, right?
Yess