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quick question - Kinetic Friction, Force Components, Blocks

 
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nate89



Joined: 28 Apr 2010
Posts: 2
Location: USA

PostPosted: Wed Apr 28, 2010 6:08 pm    Post subject: quick question - Kinetic Friction, Force Components, Blocks Reply with quote

Two blocks are stacked on a table. The upper block has a mass of 3.40 kg and the lower block has a
mass of 38.6 kg. The coefficient of kinetic friction between the lower block and the table is 0.260. The coefficient of
static friction between the blocks is 0.551. A string is attached to the lower block at 0 degrees.

what is the maxium force i can this. i know that F=MA and that Fs<Fs
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hepcj
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Joined: 23 Jun 2007
Posts: 125

PostPosted: Wed Apr 28, 2010 6:38 pm    Post subject: Reply with quote

I think that there is something wrong with the question as you have written it. Are you sure you have the figures correct? A mass of 3.4 kg on top of a mass of 38.6 kg. The first mass compared to the second is quite small so the effect of its weight would be small.
Also Fs < Fs as you have written is nonsense.
If you want an answer, then at least take the time to pose the question properly.
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nate89



Joined: 28 Apr 2010
Posts: 2
Location: USA

PostPosted: Wed Apr 28, 2010 8:11 pm    Post subject: Reply with quote

the question is right and all figures are correct.

but ill re do the question.

Two rectangular blocks are stacked on a table. The upper block has a mass of 3.40 kg and the lower block has a
mass of 38.6 kg. The coefficient of kinetic friction between the lower block and the table is 0.260. The coefficient of
static friction between the blocks is 0.551. A string is attached to the lower block and an external force −"F is applied
horizontally, pulling on the string.
(a) What is the maximum force that can be applied to the string without having the upper block slide off?
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hepcj
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Joined: 23 Jun 2007
Posts: 125

PostPosted: Fri Apr 30, 2010 12:20 pm    Post subject: Reply with quote

You can break the question down into two parts. Firstly, what acceleration is required for the top block to slip when they are both moving. Define the boundary between the block 1 and block 2 at 12 and mu the coefficient of friction between these two block as mu12.
Then the acceleration that would move the smaller block is

a= mu12 g

Next define the boundary between the larger block and the ground as 23 then mu23 is the kinetic coefficient of friction between them. To even start the blocks moving requires a force of mu23 (m1+m2) g and then an additional force is required to accelerate the blocks that would produce enough force to make the second block move. Therefore, the force required to accelerate the two blocks at this speed is:

F = mu23 (m1 + m2) g + mu12 g (m1+m2)

with the numbers you have given this works out as: 334 N
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