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A box of mass $m (kg)$ is on a slope and is attached to a wall via a light, in-extensible rod, which prevents the box from sliding down the slope. There are a few more details in the diagram above. The box is in equilibrium.

My aim is to find the value of the thrust, $T$ in the rod in terms of $m$, and the friction force, $F$ in terms of $m$, which I think we should be able to do. I don't think it matters how the metal rod is attached to the wall: it could be parallel to the slope as in the diagram, and freely hinged at the wall, for example.

In my calulations, I take $g = 9.8 \frac{m}{{sec}^2}$, and round values to $3sf$.

Resolving forces for the box parallel and perpendicular to the slope, we see the following:

$RN = mg\cos(20^\circ) = 9.21m(N)$

If the metal rod were not there, the box would accelerate down the slope with acceleration $a = 9.8 \ (\sin(20^\circ)-0.3\cos(20 ^\circ) ) = 0.589. \ $ So the resultant force down the slope without the rod would be $0.589 m(N)$.

The maximum the friction force can be is: $F = 0.3R = 0.3 \times mg\cos(20^\circ) = 2.76m(N)$, and of course this is the friction force $\iff$ the box is moving or on the point of moving.

But since the box is in equilibrium, it is not moving and so $a=0$. Resolving forces on the box parallel to the slope gets us the equation:

$F + T = mg\sin(20 ^\circ) = 3.35m(N).$

If we could make the argument that the friction force is $2.76m$ N, then that is our answer. But all we know is that $F \leq 2.76m(N)$; I don't see an obvious argument to justify equality. I don't see why, for example $F = 2.70m(N)$ and $T = 0.65m(N)$ is not correct.

  • $\begingroup$ If the box is placed against the rod, there is no reason to assume that there is any friction. $\endgroup$ – R.W. Bird Sep 18 '20 at 14:20
  • $\begingroup$ Why not? Please give a more detailed reason. $\endgroup$ – Adam Rubinson Sep 18 '20 at 14:26
  • $\begingroup$ Put it against the rod and then in contact with the surface. $\endgroup$ – R.W. Bird Sep 18 '20 at 15:04

The system as you've described it is statically indeterminate. This means that the constraints that the object is not accelerating do not provide sufficient information to uniquely determine all of the forces.

In general, one can attempt to solve statically indeterminate problems by looking at the net torque on the object as well, and setting that to be zero. However, I suspect that this will not help in this case either, since the normal force between the block and the ramp does not act at a point, and so the torque it exerts on the block is not known. If the width of the base of the block is $w$, the torque due to the normal force about the center of mass could be anything between $\pm R w/2$.

In practice, of course, "nature knows what to do". If you actually built this system, and put a force meter at the end of the rod, it would register a particular value of $T$. However, theoretically solving a statically indeterminate system generally involves taking into account the deformations that the body experiences, so knowledge of the material properties of the block would be needed. We can't simply assume that the body is perfectly rigid.

  • $\begingroup$ And before you ask: my knowledge of deformation methods to solve statically indeterminate problems is basically only "I know that such methods exist"; I don't have much knowledge of what they are or how to apply them to this situation. $\endgroup$ – Michael Seifert Sep 18 '20 at 14:58
  • $\begingroup$ This seems about right. $\endgroup$ – Adam Rubinson Sep 18 '20 at 15:11
  • $\begingroup$ @MichaelSeifert Why would there be a rotation about the center of mass of the box? Wouldn't there be a torque from the component of the box's weight to nullify that torque due to the normal force that you have talked about? $\endgroup$ – Amar30657 Sep 18 '20 at 15:25
  • $\begingroup$ @Amar30657: (a) I'm not saying that there is a rotation, I'm saying that there isn't a rotation. In statics problems we assume there is no rotation, and then use that assumption to figure out the relationships between the torques and the forces. (b) An object's weight exerts zero torque about its center of mass. In practical terms, the torques would be from the frictional force (definitely), the rod (probably, assuming its contact point isn't halfway up and it doesn't exert a shear force), and the normal force (probably, but hard to say for sure.) $\endgroup$ – Michael Seifert Sep 18 '20 at 16:33
  • $\begingroup$ @MichaelSeifert Ok, I get what you are saying but considering just for theoretical purposes that all those objects are rigid, and won't undergo deformations, there should be no frictional force on the box because the box is in rest, and the forces downward along the slope should be balanced by the contact force from the wall? You do have great reputation, could you just skim over my response? $\endgroup$ – Amar30657 Sep 18 '20 at 17:15

I would wait for some other answers, as I am a newbie here. But here it goes,

Friction will be resisting the motion created by an external force acting on an object.

Case-I: When there is no rod, your box would be moving and so the frictional force acting would be a kinetic frictional force like you have calculated.

Case-II: Now if there were a rod, your box would be stationary and now the frictional force acting would be a static frictional force. And that as my intuition dictates, will depend on the net external force on the rod+box system. In Your case, $0$ is the answer as there is no net force on that system.

The box exerts some force on the rod. There is also the weight of the box parallel to the plane of the slope on the rod. These forces are balanced by the contact force exerted by the wall to your rod. I am speculating that this force here is the third law pair for all these forces, including the thrust or the tension, that the rod has on the wall. I prefer to say there is tension instead of the thrust in the rod.

The more interesting question is what happens when you apply a force down the slope to that box.

Will there be any component of frictional force to oppose the force that we apply to that box in equilibrium, or will there always be that tension to counteract it?

I am hoping that the answer to this question is that the static frictional force will always be 0, as no matter how great a force is applied on the system of box and the rod, the wall will always be able to counteract it and thus will impede the motion of the system.

Please check my reasonings ^^

  • 2
    $\begingroup$ As far as I can tell, your speculation boils down to "if the net force on the box down the ramp is equal to the force applied by the rod up the ramp, then the force of friction will be 0." This is a correct conclusion given the premise, but it's not at all clear that the premise is true. As the OP correctly points out, Newton's Laws don't allow you to determine whether the premise is true or false. $\endgroup$ – Michael Seifert Sep 18 '20 at 19:44

I think equality can be justified. I'm new to mechanics, but I understood that once you calculated that the box would move if the rod wasn't there, you reached the conclusion that F is limiting. Then you know that F is not ≤2.76 but it is 2.76. Moreover, I'm not sure that the position of the metal rod is indifferent. If it were not parallel to the slope I think only a part of its thrust (cos(angle to the slope))would operate to stop the box, and a part of it would interfere with its weight (sin(angle to the slope))and therefore the friction.


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