Solving Masses on a Beam Problem with Scales

[delecticious]

Homework Statement

A massless beam is suspended between two scales that are separated by 6.40 m as shown in the diagram below. The scales measure a force in Newtons. Three masses are hanging from the beam: Mass X = 17.7 kg is D1 = 2.11 m from scale A, mass Y = 23.0 kg is D3 = 0.96 m from scale B (hence D2 = 3.33 m), and mass Z = 7.59 kg is hanging directly under scale B. [Use g = 9.81 m/s2.]

Homework Equations

The Attempt at a Solution

My main issue with this problem is I'm having trouble visualizing how these scales work, and since I can't visualize I can't even fathom how to approach it. Maybe it's the way the drawing is drawn but I'm not seeing how the three boxes could be weighed by scales above them, how is that even possible? Am I missing something or am I just looking at this entire picture wrong. Can anyone help me see how these scales are supposed to function?

 

[delecticious]

you know what never mind I think I just realized how I'm supposed to be looking at this. I'm supposed to user the center of gravity equation right? And if I use that how will that help me figure out what each scale reads?

 

[Doc Al]

The scales just measure the tension in the ropes they are attached to. I assume the problem is to find the scale readings. Use the conditions for equilibrium. (The sum of the forces = 0 & the sum of the torques = 0.)

 

[delecticious]

what about the tensions in the ropes connecting the boxes to the beam?

 

[Doc Al]

what about the tensions in the ropes connecting the boxes to the beam?

Those tensions just equal the weights of the boxes.

 

[delecticious]

so if I denote the one tension as T1 and the other as T2, would I be correct in saying that T1 + T2 = Wx + Wz + Wy ? or are the tensions not equal to each other?

 

[Doc Al]

so if I denote the one tension as T1 and the other as T2, would I be correct in saying that T1 + T2 = Wx + Wz + Wy ?

This is correct. Now combine this with a torque equation and you can solve for the tensions.