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Torquescrew

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Got a classic textbook physics problem here. Pretty sure I got it, but again, just wanting a "yay" or "nay" regarding if I'm doing it correctly.

And there's one little thing I'm not exactly sure of.

http://img189.imageshack.us/img189/2345/picture3pft.png

Two blocks, A and B, are placed as in Fig. 5 and con¬nected by ropes to block C. Blocks A and B weigh 20.0 N each, and the coefficient of kinetic friction between each block and the sur¬face is 0.40. Block C descends with constant velocity.

a) Draw two separate free-body diagrams showing the forces acting on A and on B.

b) Find the tension in the rope connecting blocks A and B.

c) What is the weight of block C?

Since the mass and radius of the pulleys are not given, we'll assume that they are ideal and have bearing on the equation.

basic trig

F=ma

For (a), all I had to do was doodle a bit, so let's not worry about that.

Since it's moving at a constant velocity, I know the net force has to be zero.

So, since mass A is 20 N, I just multiplied that by .4 for my force of friction.

Mu sub K A = 8 N.

For mass B, I did some fancy-pants vector smashing and wound up with a normal force of 15.9727N and a down-the-ramp force of 12.0363 N.

I multiply 15.9727 by .4 to get mu sub K B = 6.38908

So here's my first hiccup.

Friction is a two-way street (usually), so would the Fx of mass B actually be 12.0363-6.38908? (5.64722)

I'm not sure, but I don't think so.

Because ultimately mass C = 20+8+12.0363+6.38908 N, right?

Or would it be C = 20+8+5.64722+6.38908 N?

Such a small detail, but it's keeping me from finishing this assignment.

Either way, I think (but am not 100% certain) that the tension A-B will be equal to C-(12.0363+6.38908) or C-(5.64722+6.38908).

I'm right on top of the answer, but my brain has fused to the side of my skull, and I honestly can't figure out how to proceed.

And there's one little thing I'm not exactly sure of.

## Homework Statement

http://img189.imageshack.us/img189/2345/picture3pft.png

Two blocks, A and B, are placed as in Fig. 5 and con¬nected by ropes to block C. Blocks A and B weigh 20.0 N each, and the coefficient of kinetic friction between each block and the sur¬face is 0.40. Block C descends with constant velocity.

a) Draw two separate free-body diagrams showing the forces acting on A and on B.

b) Find the tension in the rope connecting blocks A and B.

c) What is the weight of block C?

Since the mass and radius of the pulleys are not given, we'll assume that they are ideal and have bearing on the equation.

## Homework Equations

basic trig

F=ma

## The Attempt at a Solution

For (a), all I had to do was doodle a bit, so let's not worry about that.

Since it's moving at a constant velocity, I know the net force has to be zero.

So, since mass A is 20 N, I just multiplied that by .4 for my force of friction.

Mu sub K A = 8 N.

For mass B, I did some fancy-pants vector smashing and wound up with a normal force of 15.9727N and a down-the-ramp force of 12.0363 N.

I multiply 15.9727 by .4 to get mu sub K B = 6.38908

So here's my first hiccup.

Friction is a two-way street (usually), so would the Fx of mass B actually be 12.0363-6.38908? (5.64722)

I'm not sure, but I don't think so.

Because ultimately mass C = 20+8+12.0363+6.38908 N, right?

Or would it be C = 20+8+5.64722+6.38908 N?

Such a small detail, but it's keeping me from finishing this assignment.

Either way, I think (but am not 100% certain) that the tension A-B will be equal to C-(12.0363+6.38908) or C-(5.64722+6.38908).

I'm right on top of the answer, but my brain has fused to the side of my skull, and I honestly can't figure out how to proceed.

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