Question about Normal Force and Torque

[Ghost Repeater]

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

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torque = Frsin(theta)

The Attempt at a Solution

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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.

 

[haruspex]

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?

 

[Chandra Prayaga]

Please show our work. How did you get a numerical answer without data on the mass of the block?

 

[Ghost Repeater]

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?

 

[haruspex]

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.

 

[CWatters]

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.