Two blocks on top of eachother with pulleys

[harrinj4]

Homework Statement

A 3.0-kg block sits on top of a 5.0-kg block which is on a horizontal surface. The 5.0-kg block is pulled to the right with a force F. The coefficient of static friction between all surfaces is 0.60 and the kinetic coefficient is 0.36.

a)What is the minimum value of F needed to move the two blocks?
b)If the force is 10% greater than your answer for (a), what is the acceleration of each block?

Homework Equations

Sum of forces = ma

The Attempt at a Solution

Forces acting on the bottom box are normal force up the y axis, normal force from the top box and mg going down the y axis. On the x axis, the F we need to find is positive and the friction of 1 on 2, the kinetic friction on the ground and the tension of the rope is going the opposite way. For the box on top, there is normal force up y axis, mg down, friction of 2 on 1 going the positive x direction and tension going the other way.

I know for the top box that n=mg so n =3(9.8) regarding the y axis
for the x axis:
F=ma
Friction of 2 on 1 = u*n = .6(3*9.8) = 17.64 so
17.64 - T = -3a


Then for the bottom box I know that :
n2-mg-n1= 0
n2 = 5(9.8)+3(9.8)
so n2=78.4

for the x direction:
F-T-kinetic friction-Friction1on2 = ma
so F-T-28.224-17.64 = 5a

And now I'm stuck!

 

[alphysicist]

Hi harrinj4,

Homework Statement

A 3.0-kg block sits on top of a 5.0-kg block which is on a horizontal surface. The 5.0-kg block is pulled to the right with a force F. The coefficient of static friction between all surfaces is 0.60 and the kinetic coefficient is 0.36.

a)What is the minimum value of F needed to move the two blocks?
b)If the force is 10% greater than your answer for (a), what is the acceleration of each block?

Homework Equations

Sum of forces = ma

The Attempt at a Solution

Forces acting on the bottom box are normal force up the y axis, normal force from the top box and mg going down the y axis. On the x axis, the F we need to find is positive and the friction of 1 on 2, the kinetic friction on the ground and the tension of the rope is going the opposite way. For the box on top, there is normal force up y axis, mg down, friction of 2 on 1 going the positive x direction and tension going the other way.

I know for the top box that n=mg so n =3(9.8) regarding the y axis
for the x axis:
F=ma
Friction of 2 on 1 = u*n = .6(3*9.8) = 17.64 so
17.64 - T = -3a

The minimum force that just starts the blocks moving, is also the maximum force that you can apply and not have the blocks move. So what is the acceleration a for the x-equations of your two blocks?

 

[harrinj4]

Hi harrinj4,



The minimum force that just starts the blocks moving, is also the maximum force that you can apply and not have the blocks move. So what is the acceleration a for the x-equations of your two blocks?

Zero?

So from there could I find T, then plug it in the other equation to find F, make acceleration zero again?

 

[alphysicist]

Zero?

So from there could I find T, then plug it in the other equation to find F, make acceleration zero again?

That's right; the acceleration is zero for both blocks in part a, so you have two equations with two unknowns (F and T) and so can solve for F.

 

[harrinj4]

That's right; the acceleration is zero for both blocks in part a, so you have two equations with two unknowns (F and T) and so can solve for F.

See I thought that, but I get 64 as the answer and it's wrong!

Hmm, see any other things I did wrong? something seems wrong...

 

[harrinj4]

This is due at five! SHITT...

 

[alphysicist]

See I thought that, but I get 64 as the answer and it's wrong!

Hmm, see any other things I did wrong? something seems wrong...

In your equation for the bottom block, you are using kinetic friction for the friction from the ground. However, it is not moving (yet), so this should be static friction.

 

[harrinj4]

In your equation for the bottom block, you are using kinetic friction for the friction from the ground. However, it is not moving (yet), so this should be static friction.

ha just figured that out a minute ago and got it right. So now for the second part:


Take the force and multiply it by 1.1 (adding ten percent), change all the frictions to kinetic frictions, and then?

 

[harrinj4]

ha just figured that out a minute ago and got it right. So now for the second part:


Take the force and multiply it by 1.1 (adding ten percent), change all the frictions to kinetic frictions, and then?

Did the work and found tension then put tension into find acceleration and it worked. Thanks for help before tho!
!

 

[alphysicist]

Did the work and found tension then put tension into find acceleration and it worked. Thanks for help before tho!
!

Perfect! (and I'm glad to help!)