What Is the Maximum Force to Prevent the Top Block from Slipping?

[tony873004]

Figure 6-36: http://www.webassign.net/walker/06-36alt.gif

5. [Walker2 6.P.071.] Two blocks, stacked one on top of the other, slide on a frictionless, horizontal surface (Figure 6-36), where M = 5.0 kg. The surface between the two blocks is rough, however, with a coefficient of static friction equal to 0.47.

If a horizontal force F is applied to the bottom block, what is the maximum value F can have before the 2.0 kg top block begins to slip?

The answer in the back of the book is 32 N. But that's not what I get. I get the same wrong answer 2 different ways:


friction = mu * mg
friction = 0.47 * 2 * 9.81
friction = 9.2214 N

at this point, I'm stuck.

I'd like to say that if pushing the 2 kg block with a force of 9.2214 will move it, then how much harder would I have to push a 5 kg block to make its push equivalent to the 2 kg push.

9.2214 * (5/2) = 23.0535 which is the wrong answer.

Or I could do it like this:

f=ma
9.2214 = 2 * a

a = 4.6107

now use this for a for the 5 kg block

F = ma
F = 5 * 4.6107
F = 23.0535 which is exactly what I got before and it is wrong

Any thoughts...??

 

[Doc Al]

You are very close. What is the maximum acceleration that the top block can have? (You already figured that out--that's when the frictional force on it is at maximum.) Now treat the two blocks as a single object (why not?). What force F is required to accelerate both blocks to that value?

(One of your mistakes was using F = ma, but not using the net force on the object. If you treat the bottom block by itself -- nothing wrong with that! -- don't forget that there are two horizontal forces on it. Recall Newton's 3rd law.)

 

[tony873004]

You are very close. What is the maximum acceleration that the top block can have? (You already figured that out--that's when the frictional force on it is at maximum.) Now treat the two blocks as a single object (why not?). What force F is required to accelerate both blocks to that value?

(One of your mistakes was using F = ma, but not using the net force on the object. If you treat the bottom block by itself -- nothing wrong with that! -- don't forget that there are two horizontal forces on it. Recall Newton's 3rd law.)

Thanks, Doc (again!)
lol... I feel so stupid for not realizing that I'm pushing 7kg, and not 5 ! I stared at this for an hour.