# Sum of torques, equilibrium problem

• Jaccobtw
In summary, the conversation involves finding the necessary force to hold up a mass, solving for x to determine the point at which force is applied, and discussing methods for calculating torque when the mass is a rod with uniform thickness and density. The conversation also includes considering the center of mass and using one end as the rotational axis for simpler algebra.
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

Last edited:
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
DEvens said:
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?

Jaccobtw said:
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.
Jaccobtw said:
Torque 3 = -(l/2 - x) * m(rod) * g
Yes.

Jaccobtw

## What is the sum of torques in an equilibrium problem?

The sum of torques in an equilibrium problem refers to the total amount of torque acting on an object that is at rest or in a state of static equilibrium. It takes into account both the magnitude and direction of the individual torques acting on the object.

## How is the sum of torques calculated?

The sum of torques is calculated by multiplying the magnitude of each individual torque by its respective lever arm and then adding all of these values together. The lever arm is the perpendicular distance from the point of rotation to the line of action of the force.

## What is the significance of the sum of torques in an equilibrium problem?

The sum of torques is significant in an equilibrium problem because it determines whether an object will remain at rest or rotate due to the forces acting on it. If the sum of torques is zero, then the object is in a state of static equilibrium and will not rotate. If the sum of torques is not zero, then the object will rotate in the direction of the net torque.

## Can the sum of torques ever be negative?

Yes, the sum of torques can be negative if the individual torques acting on the object are in opposite directions and have different magnitudes. In this case, the net torque will be the difference between the two torques, resulting in a negative value.

## What are some real-life examples of equilibrium problems involving the sum of torques?

Some real-life examples of equilibrium problems involving the sum of torques include balancing a see-saw, tightening a bolt with a wrench, and keeping a ladder in place against a wall. In all of these situations, the object is in a state of static equilibrium and the sum of torques acting on it is equal to zero.

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