# Friction and force help

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1. Nov 21, 2014

### GiantSheeps

1. The problem statement, all variables and given/known data
Three blocks in contact with each other are
pushed across a rough horizontal surface by a
98 N force as shown.
The acceleration of gravity is 9.8 m/s^2.

----F--->|2kg| |7.7kg| |9.2kg|
u = 0.082

If the coefficient of kinetic friction between
each of the blocks and the surface is 0.082,
find the magnitude of the force exerted on the
7.7 kg block by the 9.2 kg block.

2. Relevant equations
f=ma
fk=uN

3. The attempt at a solution
So i added the three masses and multiplied all of that by that acceleration due to gravity to get a normal force of 185.22
i multiplied that number by the friction to get 15.18804, and since that was the force going to the left, i subtracted that from 98 to get 82.81196

i thought that would work as a final answer because force would be consistent throughout the system, right? but i got it wrong. where did i go wrong? did i miss a step? is there something more that i need to do?

(also i tried to recreate the diagram with just typing characters, the | | are supposed to represent the blocks)

2. Nov 21, 2014

### jack action

Separate the blocks and do a free body diagram for each one, starting with the 2 kg block.

3. Nov 21, 2014

### Dick

No, force isn't a constant on all blocks. What is a constant is that each block accelerates at the same rate. Use that.

4. Nov 21, 2014

### GiantSheeps

okay so for the first block
weight is 2(9.8) = 19.6
normal force is 2(9.8) = 19.6
and friction force is .082(19.6) = 1.6072
force is 98

so 98 - 1.6072 = 96.3920 is net force

using f=ma

96.3920 = 2a
96.3920 / 2 = 48.1964

so a = 48.1964 m/s^2? that seems a bit too high, where did i go wrong? or is that actually the acceleration?

5. Nov 21, 2014

### GiantSheeps

am i even using friction correctly?

6. Nov 21, 2014

### jack action

You forgot to include the fact that the two other blocks are pushing on the first block. Both with friction and their inertia. Hint: like Dick said, acceleration is the same for all blocks.

Take the time to do the free body diagram for each block, one at a time.

7. Nov 21, 2014

### GiantSheeps

okay so taking into account what you all said i added up the masses of each block and got 18.9kg, i multiplied that number by 9.8 to get the normal force, which was 185.22 i multiplied that by .082 to get the frictional force, which got me 15.18804, then i subtracted that number from 98 to get the net force, which was 82.1196, which i divided by 18.9 to get the acceleration of the whole system, which was 4.38 and since acceleration is constant throughout the entire system, i used that number to get the net force on each block

so i got 2kg block = 8.76N
7.7 kg block = 33.738N
9.2 kg block = 40.31N

i then used all this to draw the free body diagrams, now, would i add the net force of the 7.7kg block to that of the 9.2 kg block? Or would the answer simply the net force of the 7.7kg block since every action has an equal and opposite reaction so the 7.7 kg block pushes on the 9.2 kg block with a force of 33.738N, wouldn't the 9.2 kg block push back with that same force?

(I might have missed some details when actually drawing the diagrams, free body diagrams have never really been my strong suit)

8. Nov 21, 2014

### GiantSheeps

or i just realized, perhaps i would subtract the net force of the 9.2 block by the net force of the 7.7 block to get 6.53N?

9. Nov 22, 2014

### jack action

Sorry, I had to go to bed at one point.
All good.
Similarly, the middle block has the first block pushing on one of its side (considering friction and inertia, plus the 98 N force) and the last block pushing on its other side (considering friction and inertia); The last block has the other two blocks pushing on one of its side only (considering friction and inertia, plus the 98 N force). That last sum of forces is the «magnitude of the force exerted on the 7.7 kg block by the 9.2 kg block» that you are looking for (as well as the magnitude of the force exerted on the 9.2 kg block by the 7.7 kg block).