Solving a Problem: Find Mass of Box, Given Two Wire Support Beam Data

AI Thread Summary
To find the mass of the box supported by two vertical wires, the tensions in the wires and the distances from the beam's edges are crucial. The left wire has a tension of 500 N, and the right wire has 600 N, with the box hanging 12 m from the left edge of a 16 m beam. The equilibrium condition involves balancing moments around a pivot point, which leads to the equation 8 m * 500 N + 4 m * weight of the box = 8 m * 600 N. The solution was found by correctly applying these principles, confirming that the wires are vertical and the beam is uniform. Understanding equilibrium equations is essential for solving similar problems.
pkossak
Messages
52
Reaction score
0
I really could use help on this one! I got this one wrong on the last test, it was driving me crazy. Spent almost half the time trying to figure it out.

Two wires support a beam of length L=16. A box hangs from a wire which is connected a distance of 12 m from the left edge of the beam. The tension in the left support wire is 500 N and the tension in the right support wire is 600 N. What is the mass of the box? DATA: acceleration of gravity=9.80 m/s2

My next test is today, thank you incredibly much if you can help me figure out what I was doing wrong here.
 
Physics news on Phys.org
Tell us what you did. (Are you given the mass of the beam? Is it uniform? Are the support wires vertical?)
 
Not given the mass of the beam, yes the support wires are vertical.

The figure can be seen here http://www.msu.edu/~kossakze/physics.gif

The way I approached it was to consider the middle of the beam at m = 0. The left wire is at 8 m and 8 m*500 N + 4 m*x N (weight of box) = 8 m*600 N

nevermind...wow...got the answer, guess I didn't do that on the test.

Thanks anyway
 
Good. As long as the wires are vertical and beam is uniform, no problem. Just set up your equations for equilibrium.
 
I multiplied the values first without the error limit. Got 19.38. rounded it off to 2 significant figures since the given data has 2 significant figures. So = 19. For error I used the above formula. It comes out about 1.48. Now my question is. Should I write the answer as 19±1.5 (rounding 1.48 to 2 significant figures) OR should I write it as 19±1. So in short, should the error have same number of significant figures as the mean value or should it have the same number of decimal places as...
Thread 'A cylinder connected to a hanging mass'
Let's declare that for the cylinder, mass = M = 10 kg Radius = R = 4 m For the wall and the floor, Friction coeff = ##\mu## = 0.5 For the hanging mass, mass = m = 11 kg First, we divide the force according to their respective plane (x and y thing, correct me if I'm wrong) and according to which, cylinder or the hanging mass, they're working on. Force on the hanging mass $$mg - T = ma$$ Force(Cylinder) on y $$N_f + f_w - Mg = 0$$ Force(Cylinder) on x $$T + f_f - N_w = Ma$$ There's also...
Back
Top