Beam resting on 2 pivots | Problem in Rotational Mechanics

[warhammer]

Homework Statement

A bar of length L is resting on two pivots. The weight of the bar is W (assume it is singularly concentrated at the middle) and it is in equilibrium as a result of the two pivots, both of which exert N normal reaction at their contact point. Since it is in equilibrium, the normal reaction N is equivalent to W/2. If one of the pivots is pulled from under the bar, what is the normal reaction at the moment it "just" starts to fall?

Relevant Equations

1) W=Mg
2) N+N=W (In equilibrium)
3) Torque=F*d

When one of the pivot is pulled, just at that moment a couple is formed due to the normal reaction from the existing pivot and the weight of the bar. From the assumptions given in the question, we can state that the distance between the two forces (N & W) giving rise to the couple is L/2.

Using the crude formula for Torque, we can state that L/2 is the perpendicular distance between the two forces. Hence Torque=N*(L/2) and consequently N=(2*Torque)/L

Please help me if my understanding is correct. If I'm making mistakes, please help me find those and rectify.

 

[kuruman]

Be careful how you use the term couple. The magnitudes of the forces must be equal and the directions opposite which means that the center of mass is not accelerating. Is that the case here?

 

[warhammer]

Be careful how you use the term couple. The magnitudes of the forces must be equal and the directions opposite which means that the center of mass is not accelerating. Is that the case here?

Sir I tried to assume so while it was on the verge of being pulled as asked in the question. But I don't think that's the case here. My assumption is wrong because this won't be the case when it is at the verge of being pulled and the system would topple as soon as it is released.

Upon recognising this, I have tried to formulate this alternate solution. We may take the end from where pivot is removed as the axis of rotation. Hence we will have moment of inertia about an end of the rod as ML^2/3.

We may take the torque as Mg*(L/2). Equating this with the another formula of torque which is I*α we can calculate the angular acceleration. From this we can calculate the linear acceleration "a". Finally we may do a vector addition of the forces at the said point; Mg-N=Ma. We can use the "a" value to conclusively find "N".

Is this correct sir?

 

[kuruman]

You do not specify where the pivots are relative to the bar. Can you do that or, even better, provide a picture? Also, your strategy looks OK but it would be better to show how you apply it to get the answers. In short, please show your work.

 

[warhammer]

You do not specify where the pivots are relative to the bar. Can you do that or, even better, provide a picture? Also, your strategy looks OK but it would be better to show how you apply it to get the answers. In short, please show your work.

Surely. To elucidate further, the pivots are at the end of the bar. Please find a photo highlighting the steps I've used below~

 

[kuruman]

In that case your solution is correct.

 

[warhammer]

Thank you so much for helping me out!