# Equilibrium, Torque and Friction (hard)

1. Mar 25, 2013

### hms.tech

1. The problem statement, all variables and given/known data

A uniform solid cube, of edge 1m and weight 200N, rests on a rough horizontal plane. A light rod AB
of length 1.2m is smoothly hinged at A to a point of the plane and B is in contact with a smooth face
of the cube. The rod lies in a vertical plane perpendicular to this face and passing through the centre
of mass of the cube. The rod makes an angle of 30◦ with the horizontal (see diagram). A gradually
increasing force of magnitude PN is applied vertically downwards at the mid-point of AB until the
cube is on the point of turning about the edge through D. By considering the forces acting on the cube
at this instant, find the normal contact force at B.

2. Relevant equations

sum of Forces and the sum of torque about a point are always zero if the system is in equilibrium.

3. The attempt at a solution

I don't understand the question at all. My first query : Is this system in equilibrium ?
Can i equate the Torques about "any" point to be equal to zero (specifically talking about the cube) ?

Secondly : The one thing i think would help me get started is the Free Body Diagram for this particular set up. I noticed that there is no friction between the (mass less rod) and the cube.

The first attachment was the QUESTION , as given in the book.
The second attachment is my (flawed) Free body diagram.

I am not sure about whether the forced shown in green should exist or not. Theoretically speaking, they should not ! But if i think it intuitively, the only way a force "P" acting on the rod can cause the cube to reach the point of turning over (toppling over) is that there is friction (shown in green) that would cause torque. (Am i correct) ?

Next, i am unsure where exactly would the Normal Reaction force due to the horizontal plane act ? At D or at the mid point of CD ?

I am completely confused , any help is appreciated .

2. Mar 25, 2013

### voko

The important bit is that the cube is "on the point of turning about edge D". That means there is virtually no contact between the surface and the bottom, except edge D itself. Does that help with the FBD?

3. Mar 25, 2013

### hms.tech

It Does, but i cannot accept that statement of yours without justification . In other words, i really can't see why would we make that assumption .

Referring to my query from the first post, can you explain to me how exactly can a force "P" acting ON THE ROD cause the cube to topple even though there is no friction ?

4. Mar 26, 2013

### SammyS

Staff Emeritus
There is no force present which corresponds to either green arrow. That surface is frictionless. Any force exerted on that surface is normal to the surface.

The rod is a fixed length. It's hinged at the left end, so it can't move left. If it;s rigid, so it can't bend there is no other possibility except that its right end pushes hard enough on the face of the cube, so as to make the cube tip.

Suppose the friction between the bottom of the cube and the horizontal surface was small enough so that the cube would slide rather than tip. Would you have any trouble understanding how the force P, acting through the rod could accomplish that motion?
voko's claim is absolutely valid. The cube can't slide (You do recognize that, right?), so suppose a force was exerted at the same location on the face of the cube via a horizontal rod. just as the cube is on the verge of tipping, all the weight of the block would be supported along the edge at D.

5. Mar 26, 2013

### voko

This is not my statement. This is the problem's statement. Think what "on the point of turning" really means.

It is precisely because there is no friction. This means that the reaction of the cube can only be normal. By Newton's third law, the action of the rod is equal in magnitude and opposite in direction.

If you find this somewhat indirect, here is an alternative approach. The applied force is downward. The hinge reacts. What is the vector sum of the two forces?

6. Mar 26, 2013

### hms.tech

@voko and @SammyS

I understand , because of your logical explanations, why and how the cube would tip over.

However there is one thing which i am still not clear about, and this partly answers the question put forward by voko, can the hinge apply a horizontal force in addition to a vertical force ?

I can't imagine the hinge applying any horizontal force, can it ? If so, then please explain how .

7. Mar 26, 2013

### voko

If it cannot apply any horizontal force, then it can necessarily only apply a vertical force. But what makes the vertical direction possible and the horizontal impossible for the hinge?

8. Mar 26, 2013

### hms.tech

Nothing really !

9. Mar 27, 2013

### hms.tech

I still can't seem to solve this question :

Here is what i did but failed :

since the reaction force acts through D thus it produces no moment about D.
Weight produced an anti clockwise moment about D and Normal force at (which i need to find out) , F produced a clockwise moment about D.
Hence :
200*0.5 - F*0.5 = 0
so F= 200 ... (that is not the correct answer :( )

10. Mar 27, 2013

### voko

Where do those 0.5 factors come from?

11. Mar 27, 2013

### hms.tech

That's simple : The perpendicular distance from the line of action of the forces to the Pivot D (i am taking moments about D)

12. Mar 27, 2013

### voko

The forces are applied at two different points, and at different angles with the arm. Find the distances and find the angles.

13. Mar 27, 2013

### hms.tech

I am not quite sure what u mean by that statement.

Aren't these the distances from the line of action of the force (extrapolated) and the pivot :

Shown in Green colour in the diagram .

14. Mar 27, 2013

### voko

Yes, this is also a valid approach. However, why do you think CB = 0.5?

15. Mar 27, 2013

### hms.tech

Ah, i see, i misunderstood the question (yet again) . CB = 1.2sin(30) = 0.6

16. Mar 27, 2013

### SammyS

Staff Emeritus
Yes. The system is in equilibrium.

The torque about any point is zero.

I suggest first considering a Free-Body-Diagram for the rod. (You can use the above diagram for this.) The forces exerted on the rod are:
Force P

The forces at the hinge, depicted by red arrows.

The force depicted by the left-pointing arrow at the face of the cube, which is the (reaction) force exerted by the cube on the rod.​
Finding the torque about the hinge should tell you the force exerted on the rod by the face of the cube .

That's a start.

See what you can do from there.