Question about Normal Force and Torque

In summary, the normal force and the weight of the block act at the block's center of mass. When a horizontal force is applied, the effective line of action of the normal force is displaced away from the force, and tipping occurs when it reaches the edge of the object. The torque of the normal force is zero when calculated about the bottom right corner of the block, but not when calculated about another point.
  • #1
Ghost Repeater
32
5

Homework Statement



A block of uniform density experiences a force F to the right, applied 5/3 m from the bottom of the block. The block is 2 m high and 1 m wide. Take the pivot point to be the point at the bottom right of the block. Find the value of the force that is just able to tip the box.

Homework Equations


[/B]
torque = Frsin(theta)

The Attempt at a Solution


[/B]
The force F is the only horizontal force. There are two vertical forces, the normal force and the weight of the block. I take these both to act at the block's CM, which is its geometric center. I calculate torques and get an answer of F is approximately 29.6 N.

However, the solution in the book gets 30 N, but by a very different method that doesn't seem right to me. First, it doesn't account for the normal force at all. How can this be? Don't we take the normal force to act at the CM? If that's so, then if the pivot is the bottom right corner, doesn't the normal force have a nonzero lever arm, and therefore doesn't it exert a torque?

Second, the book solution takes the lever arm for the gravitational torque to be simply .5 m, as if the gravity force were acting midway along the bottom of the block. This doesn't seem right either. Shouldn't the lever arm for the gravity force in this case be the magnitude of the vector from the pivot to the CM, scaled by the angle between that vector and the gravity vector?

My main question here is whether the normal force exerts a torque and if not, why not.

Thanks.
 
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  • #2
Ghost Repeater said:
the normal force and the weight of the block. I take these both to act at the block's CM,
You need to consider the block as it is just about to tip, or having just started. Where do you think the normal force will be?
 
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  • #3
Please show our work. How did you get a numerical answer without data on the mass of the block?
 
  • #4
haruspex said:
You need to consider the block as it is just about to tip, or having just started. Where do you think the normal force will be?

Ah, that's it. The normal force then would be applied at the pivot point and so its torque would vanish. Correct?
 
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  • #5
Ghost Repeater said:
Ah, that's it. The normal force then would be applied at the pivot point and so its torque would vanish. Correct?
Yes. More generally, as soon as a horizontal force is applied the effective line of action of the normal force is displaced away from the force. The stronger the force, the greater the displacement. Tipping occurs when it reaches the edge of the object.
 
  • #6
The torque of the normal force is zero if you calculate it about the bottom right corner of the block.

Its not zero if you calculate it about another point.

It might seem obvious to calculate it about the pivot point but you should state that's what you are doing in your working out.
 

Related to Question about Normal Force and Torque

1. What is the definition of normal force?

The normal force is the force that a surface exerts on an object that is in contact with it. It is always perpendicular to the surface and acts as a support force against the weight of an object.

2. How is normal force related to weight?

The normal force is directly related to weight. In fact, the normal force and weight are equal in magnitude when an object is at rest on a flat surface. This is because the normal force balances out the weight, keeping the object in equilibrium.

3. Can normal force be applied at an angle?

No, the normal force can only be applied perpendicular to a surface. If a force is applied at an angle, it will have components in both the vertical and horizontal directions, but only the vertical component can be considered the normal force.

4. How is torque related to normal force?

Torque is the measure of a force's ability to rotate an object around a particular axis. The normal force can contribute to torque if it is applied at a distance from the axis of rotation. This is why it is important to consider both the magnitude and direction of the normal force when calculating torque.

5. How does normal force affect an object on an inclined plane?

On an inclined plane, the normal force is usually less than the weight of the object. This is because the weight is resolved into components parallel and perpendicular to the surface, and the perpendicular component is equal to the normal force. The smaller normal force means there is a net force acting down the incline, causing the object to accelerate.

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