Nonuniform bar suspended by two cords at angles 36.9° and 53.1°

[Cyrad2]

A nonuniform bar is suspended at rest in a horizontal position by two massless cords as shown. One cord makes the angle theta1 = 36.9 with the vertical; the other makes the angle theta2 = 53.1 with the vertical. If the length L of the bar is 6.10m, compute the distance x from the left-hand end of the bar to its center of mass.

I keep getting stuck on this one. Basically all I know is that it's a statics problem, since it's at rest. So, I know that

T1 sin(theta1) + T2 sin(theta2) = mg
T1 cos(theta1) - T2 cos(theta2) = 0
where T1 and T2 are the tensions on the cords supporting the rod.

There are so many unknowns that I'm not sure where to go from here, but I've been playing around with:

T1sin(theta1)*X - T2sin(theta2)*(L-X) = 0
(Force1 * Distance1) == (Force2 * Distance2)

I would appreciate some suggestions and hints! (i'm not asking for anyone to solve it because the problem may be on my test tommarow). Thanks a bunch!

 

[member 553083]

Here's what you need to do: 1. Calculate the mass of the bar, m = T1 sin(theta1) / g, where g is the gravitational acceleration (9.81 m/s2). 2. Calculate the tension in each cord, T1 = m * g / sin(theta1) and T2 = m * g / sin(theta2). 3. Using the equation for equilibrium (T1sin(theta1)*X - T2sin(theta2)*(L-X) = 0), solve for X. Once you have the value of X, you can find the center of mass by taking the average of the two ends of the bar: x_center_of_mass = (x + L)/2.

 

[wais]

Hi there,

It looks like you're on the right track with your equations. Since the bar is at rest, we can assume that the sum of all the forces acting on it is equal to zero. This means we can set up the following equations:

T1sin(theta1) + T2sin(theta2) = mg (equation 1)
T1cos(theta1) - T2cos(theta2) = 0 (equation 2)
T1sin(theta1)*x - T2sin(theta2)*(L-x) = 0 (equation 3)

We can solve for T1 and T2 by rearranging equation 2 and substituting into equation 1:

T1cos(theta1) = T2cos(theta2) (from equation 2)
T1 = T2cos(theta2)/cos(theta1) (substituting into equation 1)
T2sin(theta2) = mg - T1sin(theta1) (from equation 1)
T2 = (mg - T1sin(theta1))/sin(theta2) (substituting into previous equation)

Now we can substitute these values back into equation 3 and solve for x:

T1sin(theta1)*x - T2sin(theta2)*(L-x) = 0 (equation 3)
T1sin(theta1)*x - ((mg - T1sin(theta1))/sin(theta2))*sin(theta2)*(L-x) = 0 (substituting in values for T1 and T2)
T1sin(theta1)*x - (mg - T1sin(theta1))*(L-x) = 0 (simplifying)
T1sin(theta1)*x - mgL + T1sin(theta1)*L - T1sin(theta1)*x = 0 (expanding brackets)
T1sin(theta1)*L = mgL (cancelling out like terms)
x = (mgL)/(T1sin(theta1)) (solving for x)

Now we can plug in the values for mg, L, T1, and theta1 to get our final answer for x. I hope this helps! Good luck on your test tomorrow.