Sum of torques, equilibrium problem

[Jaccobtw]

Homework Statement

A horizontal 2 kg rod is 2 m long. An 8 kg block is suspended from its left end, and a 4 kg block is suspended from its right end. (a) Determine the magnitude and direction of the single extra force necessary to keep the rod in mechanical equilibrium. (b) At what distance from the left end of the rod must the point of application of this force be?

Relevant Equations

Torque = perpendicular F * radius
Σ T = 0

For part a, A force must be applied so that the entire mass can be 'held up'. Therefore the necessary force must be equal to the gravitation force on all the objects:

m(rod) * 9.8 + m(LB) * 9.8 + m(RB) * 9.8 = 137 N

For part b, (This is where I'm confused) let's set the point at which the force is to be applied as the rotational axis. This will be a distance x from the left end of the rod and we will solve for x.

Torque 1 = r(1) * m(LB) * g = x * m(LB) * g (counterclockwise)
Torque 2 = r(2) + m(RB) * g = - (l-x) * m(LB) * g (clockwise)
Torque 3 = ?

The third torque should be from the mass of the rod, but I'm confused. Should we divide the mass of the rod into two parts and find the center of mass of each part and then set one of them as a clockwise torque and the other as a counterclockwise torque? Thanks for your help

 

[DEvens]

Assume the rod is of uniform thickness along its length. Then it will have a linear density of L = 1 kg / m.

So now, with the rod supported at x, the torque produced by some small portion of the rod can be calculated. You could break the rod into two parts, one length x, the other 2-x. You could then look up the torque results for rotating a rod around its end.

Or you could be brave and do the integral. Remember that Torque 1 and Torque 2 are special cases of this integral where you have assumed the blocks have zero extent. For the rod you can't do that. You will need to integrate $$\int L s ds$$ from s = -x to s = 2-x.

 

[Jaccobtw]

Assume the rod is of uniform thickness along its length. Then it will have a linear density of L = 1 kg / m.

So now, with the rod supported at x, the torque produced by some small portion of the rod can be calculated. You could break the rod into two parts, one length x, the other 2-x. You could then look up the torque results for rotating a rod around its end.

Or you could be brave and do the integral. Remember that Torque 1 and Torque 2 are special cases of this integral where you have assumed the blocks have zero extent. For the rod you can't do that. You will need to integrate $$\int L s ds$$ from s = -x to s = 2-x.

Brave it'd be indeed haha. I supposed we could also use the center of mass of the rod and use it as a clockwise torque:

Torque 3 = -(l/2 - x) * m(rod) * g

I'm wondering what you think?

 

[haruspex]

lets set the point at which the force is to be applied as the rotational axis

The algebra would be simpler using one end as the axis, preferably the end x is measured from.

Torque 3 = -(l/2 - x) * m(rod) * g

Yes.