Equilibrium of a system consisting of two bodies

[hms.tech]

Homework Statement

The attachment shows a simple diagram of a system consisting of a Sphere(uniform) and and rod(uniform) in equilibrium. The rod is in rough contact with the sphere and hence there would be friction (acting vertically at the point of contact of the two objects).

Both objects rest on a horizontal rough ground so that the friction acting on the rod is F and the frictional force acting on the sphere is -F .
Both will have the same magnitude because the system is in equilibrium.

The normal force N is also present at the point of contact.

The Attempt at a Solution

Here is the problem

Resolving horizontally for the rod only should give us the result that N=F.

Is that true or false ?

If false, then is it wrong to resolve for just one object among the two ?

 

[PhanthomJay]

Hmm well there I'm not sure of your sketch the ball will sit there just fine without the rod so I guess you mean what happens if you apply a horizontal force leftward to the sphere in which case yes the friction force at the base of the rod would equal the normal force between the sphere and top of the rod. There wouldn't be any friction force acting at the base of the sphere, though. These are ' 2 force' "members".

 

[hms.tech]

There wouldn't be any friction force acting at the base of the sphere, though. These are ' 2 force' "members".

How can you justify that statement ?

The sphere is also in equilibrium so the net force (horizontally and vertically ) on it must be zero.

The diagram shows that there is a force acting on the sphere = -N

and so the friction between the sphere and the rough ground must be equal as well .

 

[haruspex]

Resolving horizontally for the rod only should give us the result that N=F.

Yes, that's quite correct. What next? I assume there's more to the question.

 

[PhanthomJay]

How can you justify that statement ?

The sphere is also in equilibrium so the net force (horizontally and vertically ) on it must be zero.

The diagram shows that there is a force acting on the sphere = -N

and so the friction between the sphere and the rough ground must be equal as well .

No, that is not correct, as I interpret your sketch.

First of all, you should show separate sketches for the problem itself ( a sketch showing the ball and rod, and the applied load of N to the left, and the unhnown reaction loads at the base of the sphere and rod. This sketch must NOT include the force you have labeled as N to the right. This is internal to the system. You want to show just external loads (the applied force and the reaction forces).

Now before you do your FBD's, get all the reactions you can find using the equilibrium equations applied to your system. You will find that there is a downward reaction component at the rod and an upward reaction component (equal and opposite) at the sphere, which you could calculate if you knew the radius of the sphere. Now at this point there also must be horizonatl reaction components that sum to N, for equilibrium.

Now you can draw a FBD of the rod. It is a 2-force member. The horizontal component of the reaction must be equal to N, but you have shown it acting in the wrong direction. There is no horizontal reaction at the sphere, since equilibrium conditions in the x direction prohibit it.

Now if I have not interpreted your sketch correctly, please advise.

 

[haruspex]

No, that is not correct, as I interpret your sketch.

First of all, you should show separate sketches for the problem itself ( a sketch showing the ball and rod, and the applied load of N to the left, and the unhnown reaction loads at the base of the sphere and rod.

I don't understand the objections you have to the diagram as posted. It all seems reasonable to me. There are no external forces required.

 

[PhanthomJay]

If there are no external forces, there are no internal forces either. Everything is zip, as in zero. Unless the rod and sphere have weight, which is an external force; but I am assuming the weights are negligible. Please (hms.tech) state the problem as written.

 

[haruspex]

I am assuming the weights are negligible.

Since the objects 'rest' on the ground, I assumed they have weight. Then everything makes sense.

 

[Chestermiller]

This seems like a very straightforward problem. Just draw separate free body diagrams for the sphere and for the rod, showing all the forces acting on each of them. Then perform force and moment balances on the two objects. The forces N and F are equal, and will be determined by the weight of the rod, and the ratio of the rod length to the sphere radius. One thing you might consider is that the normal force between the rod and the sphere, and the tangential force will be equal. You would need a coefficient of friction greater than 1.0 to prevent slippage at the contact point. Otherwise the system would not be able to maintain equilibrium.

 

[PhanthomJay]

How can you justify that statement ?

The sphere is also in equilibrium so the net force (horizontally and vertically ) on it must be zero.

The diagram shows that there is a force acting on the sphere = -N

and so the friction between the sphere and the rough ground must be equal as well .

Ok, your first sketch should show the weight of the sphere acting down through its center, the weight of the rod acting down at its midpoint, and the vert and horiz reactions at the base of sphere and rod. You should not show the internal force N at the contact point between sphere and rod. Then you can do your FBD of the rod , and another of the sphere, and conclude that F = N. your directions of F are thus correct. Sorry I misinterpreted your sketch.