Posted by **Josh** on Tuesday, January 25, 2011 at 3:02am.

A 390 kg motor boat, moving down a frictionless ramp, is connected over a massless pulley to a 193 kg counterweight that rests on a horizontal surface. The ramp is inclined by 28.6°, and the co-efficient of kinetic friction between the couterweight and the horizontal surface is 0.507.

What is the tension in the rope connecting the boat and the counterweight?

Formula:

X-axis: m1gcos0=mTa

m2gcos9=mTa

Y-Axis: mgsin0 = 0

So I'm really stuck here. I tried to figure out the relative formula's but I think I'm doing it wrong. I just need help getting started. Thanks!

- Physics -
**bobpursley**, Tuesday, January 25, 2011 at 8:48am
It is uncertain to me where the weight is, vertical or not to the pulley. It matters.

why are these questions labeled physics? Do you have a math teacher teaching physics?

- Physics -
**drwls**, Tuesday, January 25, 2011 at 8:50am
I do not know why you have cos9 in the second formula, nor what mTa is supposed to mean. I suppose the number 0 is supposed to be theta (28.6 degrees.

There is no hyphen in coefficient.

I do appreciate your showing your work. Almost no one does that here anymore, and I wish they would.

You need to write two equations of motion: one for the direction of the boat down the ramp, and one for the counterweight in the horizontal direction. Rope tension T will be a second variable you must solve for with the two equations.

390*g sin28.6 - T = 390 a

T - 193*g*(0.507) = 193 a

You can eliminate T by adding the two equations. Once you have a, solve for T using either equation.

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