Friction Forces (Static Friction)

[huybinhs]

Homework Statement

A boy has stacked two blocks on the floor so that a 5.0 kg block is on top of a 2.0 kg block.

1. If the coefficient of static friction between the two blocks is 0.40, and the coefficient of static friction between the bottom block and the floor is 0.22, with what force horizontal should the boy push on the upper block to make both blocks in the system start to slide along the floor.

2. If he pushes too hard, the top block starts to slide off the lower block. What is the maximum force with which he can push without that happening?

Homework Equations

Fs = micro(s) * N

The Attempt at a Solution

1.

Fs (between 2 blocks) = 0.40 * (5+2) * 9.8 = 27.44 N

Fs (between bottom block and the floor) = 0.22 * 2 * 9.8 = 4.312

Fnet = 27.44 + 4.312 = 31.752 N => wrong answer

2.

Fs (maximum) = 27.44 N => wrong answer

Please help! Thanks!

 

[dineshnaveen]

yes u r correct

 

[huybinhs]

yes u r correct

How about singnificant fingure here?

 

[huybinhs]

yes u r correct

No. I'm just submited the answer, but it's INCORRECT. Please help!

 

[E=mc^84]

Homework Statement

A boy has stacked two blocks on the floor so that a 5.0 kg block is on top of a 2.0 kg block.

1. If the coefficient of static friction between the two blocks is 0.40, and the coefficient of static friction between the bottom block and the floor is 0.22, with what force horizontal should the boy push on the upper block to make both blocks in the system start to slide along the floor.

2. If he pushes too hard, the top block starts to slide off the lower block. What is the maximum force with which he can push without that happening?

Homework Equations

Fs = micro(s) * N

The Attempt at a Solution

1.

Fs (between 2 blocks) = 0.40 * (5+2) * 9.8 = 27.44 N

Fs (between bottom block and the floor) = 0.22 * 2 * 9.8 = 4.312

Fnet = 27.44 + 4.312 = 31.752 N => wrong answer

2.

Fs (maximum) = 27.44 N => wrong answer

Please help! Thanks!

This question is quite long to write out so ill briefly explain: You have to isolate the the top block and get the acceleration of just that block. Then, you use this acceleration for the bottom and the top block m1+m2 = 7kg total and you shud get the maximum force needed.

Hope this helps:)

 

[huybinhs]

This question is quite long to write out so ill briefly explain: You have to isolate the the top block and get the acceleration of just that block. Then, you use this acceleration for the bottom and the top block m1+m2 = 7kg total and you shud get the maximum force needed.

Hope this helps:)

Fs = 0.40 * 5 * 9.8 = 19.6 N

a = 19.6 / 5 = 3.92 m/s^2

so the maximum force = 3.92 * 7 = 27.44 N which is incorrect!

What am I doing incorrect here?

 

[huybinhs]

and the first question is INCORRECT too.

"Fnet = 27.44 + 4.312 = 31.752 N => wrong answer"

Please help!

 

[E=mc^84]

Fs = 0.40 * 5 * 9.8 = 19.6 N

a = 19.6 / 5 = 3.92 m/s^2

so the maximum force = 3.92 * 7 = 27.44 N which is incorrect!

What am I doing incorrect here?

You forgot to plug in friction b/w the floor = 0.22(7kg)(9.8) = 15N, therefore

Fnetx = Fa - Ffriction, where Fa = max force needed, therefore
Fa = 3.92m/s2(7kg) + 15N = 42.44N :)

 

[huybinhs]

You forgot to plug in friction b/w the floor = 0.22(7kg)(9.8) = 15N, therefore

Fnetx = Fa - Ffriction, where Fa = max force needed, therefore
Fa = 3.92m/s2(7kg) + 15N = 42.44N :)

How about the first question "what force horizontal should the boy push on the upper block to make both blocks in the system start to slide along the floor ? "

What did I do wrong here?

Thanks!

 

[huybinhs]

Just submited the 42.44 N for the second question and it's INCORRECT. I'm confused...

 

[E=mc^84]

Homework Statement

A boy has stacked two blocks on the floor so that a 5.0 kg block is on top of a 2.0 kg block.

1. If the coefficient of static friction between the two blocks is 0.40, and the coefficient of static friction between the bottom block and the floor is 0.22, with what force horizontal should the boy push on the upper block to make both blocks in the system start to slide along the floor.

2. If he pushes too hard, the top block starts to slide off the lower block. What is the maximum force with which he can push without that happening?

Homework Equations

Fs = micro(s) * N

The Attempt at a Solution

1.

Fs (between 2 blocks) = 0.40 * (5+2) * 9.8 = 27.44 N

Fs (between bottom block and the floor) = 0.22 * 2 * 9.8 = 4.312

Fnet = 27.44 + 4.312 = 31.752 N => wrong answer

2.

Fs (maximum) = 27.44 N => wrong answer

Please help! Thanks!

Sorry , but that solution is if the bottom block is moved without the top block moving, my mistake. I am no expert at this, just trying to help. Ya, so in 1) If you want to get the system moving by just pushing on the top block then the force needed for that is just the force of friction of the two masses: 0.22(7kg)(9.8) = 15N, since they will move in equilibrium with each other.

For 2) He does not want the top block to move, so you isolate the top block to get the force of friction = 0.4(5kg)(9.8) = 19.6N, which is the max force since beyond this force you will accelerate the top block.

Hope this helps:)