# Basic forces and rotation problem

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## Homework Statement

A uniform bar of mass m is being moved on a smooth horizontal plane by applying a constant horizontal force F acting at the lowest point. If the rod translates making a constant angle 'theta' with vertical, the value of F must be?

## Homework Equations

1) F=ma
2) Torque=Moment of Inertia X Angular acceleration

## The Attempt at a Solution

I tried to equate net torque about center of rod and lowest point of rod to zero but can't figure out how to do so in either configuration, I figured the torque of the force about the center is F(L/2)cos(theta) assuming length of rod as L, I don't know how to proceed or how this will give us the value of F in terms of m, g and theta

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Have you drawn the normal force N?
You should write out the balance of vertical forces to find N.
Then choose the centre of mass as the pivot point of torques and write out the torque balance of the force F and N.

Gold Member
Have you drawn the normal force N?
You should write out the balance of vertical forces to find N.
Then choose the centre of mass as the pivot point of torques and write out the torque balance of the force F and N.
Thanks a lot, on doing this I got the correct answer as F=mgtan(theta); but I still don't understand one point- the torque about any point on the rod should be zero, but when I choose the lowest point as pivot point, F and Normal N have no torque, the only torque is its weight which gives a positive torque, what am I missing?

First notice that the only horizontal force is the force F, so the bar will have horizontal acceleration. If you choose the pivot point somewhere then it's like you choose the frame of reference of that point, which accelerates.
Now if you choose an accelerating frame of reference then you have to deal noninertial forces. This force equils m*a, where m is the mass, and a is the acceloration, its direction is opposite the direction of vector 'a' and it acts on the centre of mass.
This is the force you missing

Gold Member
First notice that the only horizontal force is the force F, so the bar will have horizontal acceleration. If you choose the pivot point somewhere then it's like you choose the frame of reference of that point, which accelerates.
Now if you choose an accelerating frame of reference then you have to deal noninertial forces. This force equils m*a, where m is the mass, and a is the acceloration, its direction is opposite the direction of vector 'a' and it acts on the centre of mass.
This is the force you missing
Alright, thanks a lot for clearing my doubts, I appreciate it