Blocks, incline, tension and acceleration

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Homework Help Overview

The discussion revolves around a physics problem involving a large block on an incline and a smaller block connected by a string over a pulley. The problem includes parameters such as mass, angle of incline, and coefficients of friction, and asks for the acceleration of the smaller block and the tension in the string.

Discussion Character

  • Mixed

Approaches and Questions Raised

  • Participants explore the equations of motion and forces acting on both blocks. There is an attempt to confirm the correctness of the original poster's calculations and reasoning regarding the direction of acceleration.

Discussion Status

Some participants question the feasibility of the scenario presented in the problem, suggesting that the hanging mass may not be sufficient to cause the block to slide up the incline. Others express uncertainty about the problem's setup and the assumptions made regarding forces and friction.

Contextual Notes

There is a noted discrepancy regarding the problem's wording and the physical plausibility of the situation, with participants discussing the implications of static and kinetic friction in the context of the problem.

noob314
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Homework Statement


A large block of mass M = 6kg is on an incline which is at an angle of 30 degrees above the horizontal. The coefficient of kinetic friction between the block and the incline is 0.3. The block is attached to a string which runs over a pulley and is connected to a smaller block of mass m = 2.4 kg.

If the block is sliding up the incline, find a) the acceleration (magnitude and direction) of the smaller block m and b) the tension in the string.


The Attempt at a Solution


I need someone to confirm if what I'm doing is correct.

a)
\Sigma \vec{F}_{M} = Ma = T - f_{k} - Mgsin\theta
T = M(gsin\theta + a + \mu_{k}gcos\theta)
\Sigma\vec{F}_{m} = ma = mg - T
= mg - M(gsin\theta + a + \mu_{k}gcos\theta)
= ma + Ma = mg - Mgsin\theta - \mu_{k}Mgcos\theta
= a(m+M) = mg - Mgsin\theta - \mu_{k}Mgcos\theta
= a = \frac{mg - Mgsin\theta - \mu_{k}Mgcos\theta}{m+M}
= a = -2.52\frac{m}{s^{2}}
I'm guessing this is the acceleration in the y-direction, and the acceleration in the x-direction would be 0, so the magnitude is 2.52
= 2.52\frac{m}{s^{2}} down

b)
\Sigma\vec{F}_{m} = mg - T = ma
T = mg - ma
= (2.4kg)(9.80 \frac{m}{s^{2}}) - (2.4kg)(2.52\frac{m}{s^{2}})
= 17.47N

Which looks about right, since in order for m to be moving down, mg needs to be greater than T, which in this case it is.
 

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I haven't checked your equations, but you should first check the wording of the problem. The block can't be sliding up the plane, the hanging mass is too small to haul it up. It can't be sliding down the plane, either. Did you copy the problem down correctly?
 
Yes, the problem is copied down correctly.
 
noob314 said:
Yes, the problem is copied down correctly.
Then someone made a boo-boo. The problem makes no sense as written.
 
Is it really impossible? I assumed some unknown force allowed it to overcome whatever static friction there may have been which would have allowed mass m to overcome the kinetic friction.
 
noob314 said:
Is it really impossible? I assumed some unknown force allowed it to overcome whatever static friction there may have been which would have allowed mass m to overcome the kinetic friction.
What unknown force?
 
I assume it would be a force that's barely able to overcome the static friction and then disappear after that.
 
The more you try to make sense of this problem, the more it will confuse you. Just rip it up and try another problem.
 
Yeah, you're right. I rechecked my calculations and the numbers didn't add up. I was just curious because it was on one of my old exams.
 

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