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HallsofIvy

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F

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I also know the force of friction is equal to the coefficient of friction times the normal force. I think the normal force is mass times the sine of theta times gravity, but I don't know how to find the force of friction.

You almost said it! The normal force is mg cos(35) (not just mass times sin- mg times cosine-look at your diagram) and the friction force is the coefficient of friction times that. The problem is that, here, you are not given the coefficient of friction- that's the "unknown" you are asked to find. Let's call it "c". The friction force on the 8 kg mass is 8(9.5)cos(35) and for the 3.5 kg mass it is 3.5(8.9)cos(35).

The "S" component of gravitational force on the 8 kg mass is pulling it down the incline, the "S" component of gravitational force on the 3.5 kg mass is holding it back (through the rope) and both frictional forces are holding it back. Taking the positive direction down the slope, we have for the total force

F= 8(9.8)sin(35)- 3.5(9.8)sin(35)- 8(9.8)c cos(35)- 3.5(9.8)c sin(35)

= (8-3.5)(9.8)sin(35)- (8-3.5)(9.8)c cos(35)

= 4.5(9.8)sin(35)- 4.5(9.8)c cos(35).

We are told that the 8 kg mass has acceleration 1.5 m/s

Solve that equation for c.

Of the four forces, two (the "S" component of gravity on the 8 kg mass and the friction force of the 8 kg mass) are acting directly on the 8 kg mass. The other two (the "S" component of gravity on the 3.5 kg mass and the friction force on the 3.5 kg mass) have to act through the string connecting the two mass. The total of those two forces is the tension in the string.

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In the book, the coefficient of kinetic friction is .0871 and tension is equal to 27.4 N.

When I try to get tension using the "S" component of gravity on the 3.5 kg mass and the friction force on the 3.5 kg mass, I figured the "S" component to be (3.5)(9.8)cos(35). I wasn't sure about the friction force, though. The friction force calculated above was for the whole system .. can still take (4.5)(9.8)sin(35)-(4.5)(9.8)cos(35) to get it?

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