Static Pulley Problem on an Incline

[equinom]

The problem is: Two blocks of mass m and M are connected via pulley with a configuration as shown. The coefficient of static friction is 0.64, between blocks and surface. If the inclined angle is 23deg and M = 3.1kg, what is the maximum mass m so that no sliding occurs?

The question is: I understand how m=0.64(3.1)/(sin23-0.64cos23) but how can I explain the negative mass? Also, when you look at the denominator there is a restriction that theta can't equal 33 degrees. What exactly does that mean?

 

[Dr.D]

You seem to have an error somewhere in your analysis.

Draw a free body diagram for each block (include ALL forces on each block). Then write the equations of motion and set the system acceleration to zero.

 

[Charles Link]

@equinom I think you calculated the angle that the denominator is less than zero incorrectly. [Editing after further analysis: See below=your calculations are correct]. That angle is around 10 degrees,(I will compute it more precisely momentarily) and what that means is the mass m will stay on the surface without creating any tension in the rope, and won't need the full available frictional force of $\mu \, mg \cos{\theta}$ to keep it from sliding. Instead the frictional force on block $m$ will be only $mg \sin{\theta}$ for angles $\theta$ of 10 degrees and less. $\\$ Edit: Scratch that=your calculation of 33 degrees is correct. What that means is at 23 degrees, its own frictional force, regardless of the value of mass $m,$ will hold the mass $m$ in place. Mass $m$ can be as heavy as you like at 23 degrees. Tension in the rope occurs only for $\theta>33$ degrees. The formula you have for the maximum mass $m$ only applies for angles $\theta>33$ degrees. $\\$ For $\theta<33$ degrees, there is no limit to the maximum $m$ that you could put there. It would be held in place by its own friction, and wouldn't need the mass $M$ to help hold it there.

 

[Charles Link]

You seem to have an error somewhere in your analysis.

Draw a free body diagram for each block (include ALL forces on each block). Then write the equations of motion and set the system acceleration to zero.

@Dr.D Please see my edited comments of post 3. I think the OP has stumbled on an interesting case where the angle of inclination is sufficiently small, that no sliding occurs, regardless of the value of the mass $m$. :)

 

[haruspex]

when you look at the denominator there is a restriction that theta can't equal 33 degrees. What exactly does that mean?

It tells you you have made a wrong assumption in obtaining the equation. You did indeed make an assumption... what was it?

 

[Dr.D]

Please see my edited comments of post 3. I think the OP has stumbled on an interesting case where the angle of inclination is sufficiently small, that no sliding occurs, regardless of the value of the mass m m . :)

While this is certainly possible (I have not calculated it), FBDs for the two bodies are still the right place to start.

 

[equinom]

"For theta<33 degrees, there is no limit to the maximum $m$ that you could put there. It would be held in place by its own friction, and wouldn't need the mass M to help hold it there.

Hmm, in an extreme case that m=1000kg, how can the friction created by a M=3.1kg block hold it in place. Something doesn't add up? Otherwise, thank-you for your explanation, it definitely makes me think about the problem a little differently.

 

[Charles Link]

Hmm, in an extreme case that m=1000kg, how can the friction created by a M=3.1kg block hold it in place. Something doesn't add up? Otherwise, thank-you for your explanation, it definitely makes me think about the problem a little differently.

@haruspex has a good question for you @equinom in post 5.

 

[equinom]

@haruspex has a good question for you @equinom in post 5.

Next question is, what is the assumption that I made? Probably that the blocks would move down the incline and that friction would be in the opposite direction. After playing with the equation a bit I get m=μM/(sinθ-μcosθ) and mathematically sinθ-μcosθ≠0 or tanθ≠μ. Is that always true or did I make an incorrect of assumption?

 

[Charles Link]

Next question is, what is the assumption that I made? Probably that the blocks would move down the incline and that friction would be in the opposite direction. After playing with the equation a bit I get m=μM/(sinθ-μcosθ) and mathematically sinθ-μcosθ≠0 or tanθ≠μ. Is that always true or did I make an incorrect of assumption?

The incorrect assumption is that the frictional force is always $F_f= \mu \, mg \cos{\theta}$. When $\theta=0$, the object will rest there freely. Is the frictional force=$\mu \, mg$ for this case? What is the frictional force for $\theta=0$, even though $\cos(0)=1$? (Also see my post 3 again).