What are the tension and angle of support rope for a cylinder about to slip?

[Vitani11]

Homework Statement

A uniform cylinder with mass 100kg, length 4.0m, and radius 12cm is supported by a rope as shown. The coefficient of static friction between the cylinder and the horizontal is 0.60. The cylinder is just about to slip to the right. What is the tension in the support rope and the angle the rope makes with the vertical wall?

Homework Equations

Sum of forces
Sum of torques

The Attempt at a Solution

Will post below.

 

[Vitani11]

I'm getting that the tension is equal to (mass x gravity) / 2

 

[TomHart]

I did not get the same result as you did. But I have to say that I don't have a lot of confidence in my result. It would help if you could show your work.

Edit: I guess I didn't answer your question in the title of the post. The FBD looks right, although I probably would have drawn the friction force facing the opposite direction.

 

[Buffu]

Homework Statement

A uniform cylinder with mass 100kg, length 4.0m, and radius 12cm is supported by a rope as shown. The coefficient of static friction between the cylinder and the horizontal is 0.60. The cylinder is just about to slip to the right. What is the tension in the support rope and the angle the rope makes with the vertical wall?...

Are you sure of direction of frictional force ?

 

[Vitani11]

That's how the problem was stated. I didn't think it would be physically possible to slip right either, if that's what you're saying.

 

[TSny]

I didn't think it would be physically possible to slip right either, if that's what you're saying.

Note that the tension force has a horizontal component to the right. What keeps the cylinder from sliding to the right?

Buffu is asking you to think about the direction of the friction force acting on the cylinder.

 

[haruspex]

That's how the problem was stated. I didn't think it would be physically possible to slip right either, if that's what you're saying.

Drawing the force as acting to the right only means that is taken as the positive direction for purposes of writing equations. It does not matter whether it will in fact act that way. If it doesn't, the value will come out negative. You just have to be consistent in the algebra.

 

[TSny]

Drawing the force as acting to the right only means that is taken as the positive direction for purposes of writing equations. It does not matter whether it will in fact act that way. If it doesn't, the value will come out negative. You just have to be consistent in the algebra.

Often a negative answer for an unknown force tells you that the direction of the force is opposite to what you assumed, but the magnitude is nevertheless correct.

But when dealing with friction you sometimes have to be careful. If I set this problem up with friction assumed to be toward the right, then I get very different answers for the magnitudes of N, f, T and θ compared to assuming friction is toward the left.

 

[haruspex]

Often you can use the sign of the answer for an unknown force to tell you that the direction of the force is opposite to what you assumed, but the magnitude is nevertheless correct.

But when dealing with friction you sometimes have to be careful. If I set this problem up with friction assumed to be toward the right, then I get very different answers for the magnitudes of N, f, T and θ compared to assuming friction is toward the left.

Yes and no. In general, static friction force is constrained by $|F_{sf}|<=\mu_s|F_{normal}|$. If told it is on the point of slipping that becomes $|F_{sf}|=\mu_s|F_{normal}|$. The correct sign should emerge from consideration of the horizontal force balance.

 

[TSny]

Yes and no. In general, static friction force is constrained by $|F_{sf}|<=\mu_s|F_{normal}|$. If told it is on the point of slipping that becomes $|F_{sf}|=\mu_s|F_{normal}|$. The correct sign should emerge from consideration of the horizontal force balance.

Yes. But if I blindly assume the wrong direction for the friction force, substitute |f_s| = μ_s |N|, and then just mechanically go through the algebra, I get different results for the magnitudes of the forces compared to if I had assumed the correct direction for the friction force.

As you say, you can deduce the correct direction of the friction force from the balancing of forces, as I was hinting to the OP in post #6.

 

[haruspex]

if I blindly assume the wrong direction for the friction force, substitute |f_s| = μ_s|N|

No, it's only a problem if you substitute f_s = μ_s|N|.

 

[TSny]

No, it's only a problem if you substitute f_s = μ_s|N|.

Well, maybe we can come back to this later. I don't want to write any equations at this point since Vitani11 has yet to show us any equations.

 

[Vitani11]

Here is what I have. The direction for the first term in the torque equation should be negative, but it's fixed in the next line. I used the base of the cylinder as the pivot.

 

[haruspex]

Here is what I have. The direction for the first term in the torque equation should be negative, but it's fixed in the next line. I used the base of the cylinder as the pivot.

Your torque equation looks wrong in several ways. You are taking moments about the point of contact with the ground, right?
What is the horizontal distance from that point to the mass centre of the cylinder (be careful)?
If you resolve the tension into horizontal and vertical components, each has a moment about the axis.

 

[SammyS]

The line extending from the point of contact of the cylinder with the floor and through the center of mass, continuing on through point of attachment of the rope, does not make an angle of 20° with the horizontal. That angle is somewhat more than 3° (actually about 0.06 rad) greater than 20°.

 

[Vitani11]

I've gotten the torque equation as τ= -(L/2)mgsinϕ - LsinϕTsinθ + LTcosθcosϕ. L is the length of the cylinder and T is tension.

 

[haruspex]

I've gotten the torque equation as τ= -(L/2)mgsinϕ - LsinϕTsinθ + LTcosθcosϕ. L is the length of the cylinder and T is tension.

I don't know whether your φ is the 20 degrees or the 70 degrees, but either way, the horizontal distance from point of contact to mass centre is not (L/2)sinϕ, the horizontal distance to point of attachment of cable is not Lcos φ, and the vertical distance to point of attachment is not L sin φ. The cylinder has width.

 

[SammyS]

I've gotten the torque equation as τ= -(L/2)mgsinϕ - LsinϕTsinθ + LTcosθcosϕ. L is the length of the cylinder and T is tension.

However, $\ \phi \ne 20^\circ \$ .

 

[haruspex]

However, $\ \phi \ne 20^\circ \$ .

But is not sufficient to change the angle. The length has to be changed too. I feel it is simpler not to worry about calculating the right length and the right angle, but simply work in terms of components of distance parallel to and normal to the cylinder's axis,