Minimum and maximum values so the system remains at rest

[James2911]

Homework Statement

[/B]
Two blocks A and B of masses m and M are connected to the two ends of a string passing over a pulley. B lies on plane inclined at angle theta with the horizontal and A is hanging freely as shown. The coefficient of static friction between B and the plane is us. Find the minimum and maximum values of m so that the system is at rest.

In the given diagram: B (i.e. M) lies on the incline plane
A (i.e. m) is hanging
Theta is the angle of the incline
Both A and B are connected by a pulley.

Homework Equations

Sigma F = ma[/B]

The Attempt at a Solution

Given[/B]

system is at rest;
for A; mg = T -- (1)
for B; Mgsin(theta) + uMgcos(theta) = T

from (1)
=> Mgsin(theta) + uMgcos(theta) = mg
=> Msin(theta) + uMcos(theta) = m

How do I proceed from here? Do I need to differentiate?

 

[Doc Al]

You actually solved the problem for one case. Which one? Is the one you solved the max value of m or the minimum?

How do the two cases differ?

 

[Ray Vickson]

Homework Statement

[/B]
Two blocks A and B of masses m and M are connected to the two ends of a string passing over a pulley. B lies on plane inclined at angle theta with the horizontal and A is hanging freely as shown. The coefficient of static friction between B and the plane is us. Find the minimum and maximum values of m so that the system is at rest.

In the given diagram: B (i.e. M) lies on the incline plane
A (i.e. m) is hanging
Theta is the angle of the incline
Both A and B are connected by a pulley.

Homework Equations

Sigma F = ma[/B]

The Attempt at a Solution

Given[/B]

system is at rest;
for A; mg = T -- (1)
for B; Mgsin(theta) + uMgcos(theta) = T

from (1)
=> Mgsin(theta) + uMgcos(theta) = mg
=> Msin(theta) + uMcos(theta) = m

How do I proceed from here? Do I need to differentiate?

If $u$ and $\theta$ are specified, your equation gives you $m$ in terms of $M,u$ and $\theta$, so you have no opportunity to maximize or minimize or anything else. As far as I can see, you only have a chance to maximize or minimize if you are allowed to vary $\theta$.

 

[Doc Al]

Do I need to differentiate?

No.

 

[James2911]

You actually solved the problem for one case. Which one? Is the one you solved the max value of m or the minimum?

How do the two cases differ?

This is where my doubt lies. How do I determine which one is which?
You said I solved the problem for one case, but what I am thinking is that I have only arrived at the relation and from here I need to manipulate it in order to arrive at the required answer.

If $u$ and $\theta$ are specified, your equation gives you $m$ in terms of $M,u$ and $\theta$, so you have no opportunity to maximize or minimize or anything else. As far as I can see, you only have a chance to maximize or minimize if you are allowed to vary $\theta$.

u is μ (coefficient of friction)
and θ is the angle of the inclined plane.

I think θ remains fixedThe given answer is:
m_min = M(sinθ - μ cos θ)
m_max = M(sinθ + μcosθ)

 

[Doc Al]

This is where my doubt lies. How do I determine which one is which?

Here's a hint: When m is below the minimum, which way will M tend to slide? Same question for when m is above the maximum. (Consider the direction of the friction force in each case.)

 

[Ray Vickson]

This is where my doubt lies. How do I determine which one is which?
You said I solved the problem for one case, but what I am thinking is that I have only arrived at the relation and from here I need to manipulate it in order to arrive at the required answer.

u is μ (coefficient of friction)
and θ is the angle of the inclined plane.

I think θ remains fixedThe given answer is:
m_min = M(sinθ - μ cos θ)
m_max = M(sinθ + μcosθ)

Your last equation in post #1 says

Msin(theta) + uMcos(theta) = m

so given $M, u$ and $\theta,$ $m$ is uniquely determined. There is no opportunity to maximize or minimize; all you can do is compute $m.$

The only other possibility is that you have somehow written incorrect equations.

 

[Doc Al]

Msin(theta) + uMcos(theta) = m

so given $M, u$ and $\theta,$ $m$ is uniquely determined. There is no opportunity to maximize or minimize; all you can do is compute $m.$

And that value will be one of the answers. Note that the equations change for each case.

 

[Ray Vickson]

And that value will be one of the answers. Note that the equations change for each case.

Right, and I think the OP needs to draw a complete free-body-diagram, showing all the forces (gravity, friction, tension, etc). That will show the nature of the system when it is stuck in place, versus when it is just about to start sliding, one way or the other.

Personally, I think a useful way to think about it is to look first at the condition for equilibrium when there is no friction. That will give a starting value $m_0$, so that if $m < m_0$ then $M$ will start to slide one way and if $m > m_0$ it will start to slide the other way----all in the absence of friction. Now add friction and see what happens.

 

[ehild]

Given

system is at rest;
for A; mg = T -- (1)
for B; Mgsin(theta) + uMgcos(theta) = T

Remember, the static friction force is not uMgcos(theta), it is less or equal to it. So you need to write inequalities for block B, one to preventing acceleration downward, second to prevent acceleration upward along the slope. The friction opposes the motion, so it is different for the two cases.

 

[James2911]

Here's a hint: When m is below the minimum, which way will M tend to slide? Same question for when m is above the maximum. (Consider the direction of the friction force in each case.)

This is it! I did what you said, and I got the exact answers!

When m is below the minimum, M will slide down
When m is above the maximum, M will slide up
&
Friction will orient itself according to motion. (Opposing motion in each case)

I now see why the answer in my original post was one of the answers. I had assumed initially that m will go down and M will slide up.

Thank you Doc Al, Ray Vickson and ehild for your help!

 

[Doc Al]

Good work!