- #1

Jaccobtw

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- 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

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

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