Solve Mass and Dist. of Bar Weights: x-m

[mattmannmf]

A beam of mass mb = 10.0 kg, is suspended from the ceiling by a single rope. It has a mass of m2 = 40.0 kg attached at one end and an unknown mass m1 attached at the other. The beam has a length of L = 3 m, it is in static equilibrium, and it is horizontal, as shown in the figure above. The tension in the rope is T = 637 N.

a) Determine the distance, x, from the left end of the beam to the point where the rope is attached. Note: take the torque about the left end of the beam.

I know that M1 is 15 kg. I keep getting 4 m from the point of rotation but that's wrong. here is the help they told me, but i just don't think it makes sense at all:

HELP: We know the sum of the torques will be zero. Since we are taking the torque about the left end of the beam, there are only three torques we will need to deal with.

HELP: We can assume the beam is uniform in distribution of mass and therefore take the torque of the beam at the center, 1.5 meters. Remember that the torque is the force time the perpendicular distance, i.e. be careful with the distances you choose. The torque is zero about every point, and therefore you can take the torque about another point as a check.

 

[hage567]

It would be helpful if you showed your calculations. We have no way of knowing what you did otherwise.

 

[nlightner]

First, let's clarify some of the given information. The beam has a total mass of 10 kg and a length of 3 m. This means that the mass is evenly distributed throughout the beam, and we can assume that the center of mass is at the midpoint of the beam, which is at x=1.5 m from the left end. This is important because we will be using this point as our reference for calculating torques.

Now, let's look at the three torques that we need to consider. The first is the torque due to the weight of the beam itself, which is acting at its center of mass. The second is the torque due to the weight of m2, which is acting at a distance of 3m from the left end (since it is attached to the end of the beam). And finally, the third torque is due to the tension in the rope, which is acting at a distance of x from the left end.

Since the beam is in static equilibrium, the sum of these three torques must be equal to zero. This gives us the following equation:

Στ = 0

Where Στ is the sum of all the torques. Now, we can write out the individual torques:

τ1 = Wbeam * d1 = (10 kg * 9.8 m/s^2) * (1.5 m) = 147 Nm

τ2 = Wm2 * d2 = (40 kg * 9.8 m/s^2) * (3 m) = 1176 Nm

τ3 = T * d3 = (637 N) * (x m)

Where d1, d2, and d3 are the perpendicular distances from each torque to our reference point (the center of the beam).

Now, we can plug these values into our equation and solve for x:

Στ = τ1 + τ2 + τ3 = 0

147 Nm + 1176 Nm + (637 N * x m) = 0

637 N * x m = -1323 Nm

x = -1323 Nm / 637 N = -2.08 m

This answer may seem strange, but remember that we are taking the torque about the left end of the beam, so the negative sign simply means that the distance is to the left of our reference point. Therefore, the actual distance from