Magnitude of the tension when there is static (variable) friction

[arestes]

Hello folks,
I'm being tricked by the pesky friction force again.
I know that static friction acts just to put everything in equilibrium as long as its magnitude doesn't exceed a certain value. That is ok. But when there are other adjustable force such as TENSION and NORMAL, (which in this case DOESN'T have to be related to the friction by f=\mu \times normal) , how do these three reach an agreement?

To put an example. I modified a figure (attached here in this post) to illustrate this: We have a rope attached to a wall on one end and attached to a box on the other. The box is resting on an incline with angle theta that has a certain coefficient of static friction. Let's not bother about the value, I'll just assume the friction never exceeds its allowed maximum. There are three unknowns: normal, tension and friction. Granted, there is an inequality for the friction (shouldn't exceed mu times normal) but that isn't helpful to solve the system. However, we can intuitively see that this system HAS to be determinate.

the equations I get, assuming that friction goes down the slope (direction could be fixed if this were solvable and we looked at the sign) are:
-T sin(\theta) - mg cos (\theta) + n= 0 and
-f+Tcos(\theta) -mg sin(\theta) = 0

\theta, and m are given, so we have three unknowns and two equations. This is not an extended body so torques can't possibly help. What am I missing?

I know some textbook problems give an additional (all-important) piece of data: the body is just about to slide down or go uphill, but in this case, I think the tension cannot be adjusted to have two cases, this seems intuitively to have a unique case.

BTW, this is no homework problem. I already graduated some years ago... just not practicing in a while. Thanks for any help.

 

[Chestermiller]

The tension is not an unknown in this system. It is imposed. The normal force and shear force are functions of the imposed tension.

 

[arestes]

hi, thanks. But that's exactly what I can't imagine in real life. How can we impose or fix the tension if we just attach it to the wall? This seems to me like it would naturally find an equilibrium. Maybe you could make me realize where or how an external adjustment is needed or inadvertently done?
Is this one of those cases seen in engineering where elastic properties of the materials have to be taken into account to solve an statically indeterminate system?
thanks

 

[Chestermiller]

hi, thanks. But that's exactly what I can't imagine in real life. How can we impose or fix the tension if we just attach it to the wall? This seems to me like it would naturally find an equilibrium. Maybe you could make me realize where or how an external adjustment is needed or inadvertently done?
Is this one of those cases seen in engineering where elastic properties of the materials have to be taken into account to solve an statically indeterminate system?
thanks

Not necessarily. Suppose we start out by just leaving the rope slack. As long as the box is not sliding, we know the normal force and the tangential force, and we know that the tangential force on the box is up the incline. Now we start to impose tension on the rope. As we do so, the box still doesn't move, but the tangential force up the incline decreases. At some point, the tension on the rope will be high enough so that the tangential force up the incline will be zero. But now, we continue to increase the tension in the rope. Now, however, the tangential force on the box will be up the incline.

If we assume that the rope is inextensible, then the problem would have to be regarded as statically indeterminate. But, if the rope is extensible, then we could achieve the desired tension by shortening up on the section of rope between the box and the wall, while maintaining the distance between the box and the wall constant.

 

[SteamKing]

I can't see where your problem is statically indeterminate. I think your first order of business is to convert your sketch into a free body diagram.

If the weight of the box and the angle of the ramp are known, then the weight is resolved into the normal force N and another component which acts parallel to the surface of the ramp. From the normal force and the coefficient of friction, then the magnitude of the friction force F can be determined. The friction force always acts to oppose motion, so in this case, it would be pointed up the ramp. Once the magnitudes of these forces are determined, then the magnitude of the tension in the rope must be that which keeps the box from moving down the ramp, that is, the tension keeps the box in equilibrium with respect to the ramp.

Obviously, the magnitude of the tension will be a maximum if there is no friction between the ramp and the box. If the box were stuck to the ramp, then no tension would be required to keep the box in equilibrium.