 5
 1
1. Homework Statement
Consider a rigid 2.60mlong beam (see figure) that is supported by a fixed 1.30mhigh post through its center and pivots on a frictionless bearing at its center atop the vertical 1.30mhigh post. One end of the beam is connected to the floor by a spring that has a force constant k = 1100 N/m. When the beam is horizontal, the spring is vertical and unstressed. If an object is hung from the opposite end of the beam, the beam settles into an equilibrium position where it makes an angle of 22.6° with the horizontal. What is the mass of the object?
2. Homework Equations
∑F_{x}=0
∑F_{y}=0
∑torque = 0
3. The Attempt at a Solution
I don't have any forces working horizontally, and I don't think I need to set up an equation for the vertical forces because I don't have the mass of any of the beams. So that leaves me to set up a torque equation:
mgcos(22.6)(L/2) = kx(L/2)
m = (kx)/(gcos(22.6))
I need to solve for x, and when I originally attempted this problem I solved for x like this:
x = (L/2)tan(22.6)
But I didn't get the correct answer for m with this x, and I think it is because I solved for x as if it was perpendicular to the dashed line in the diagram. I am currently stuck trying to find x(change in length of spring from rest)
Consider a rigid 2.60mlong beam (see figure) that is supported by a fixed 1.30mhigh post through its center and pivots on a frictionless bearing at its center atop the vertical 1.30mhigh post. One end of the beam is connected to the floor by a spring that has a force constant k = 1100 N/m. When the beam is horizontal, the spring is vertical and unstressed. If an object is hung from the opposite end of the beam, the beam settles into an equilibrium position where it makes an angle of 22.6° with the horizontal. What is the mass of the object?
2. Homework Equations
∑F_{x}=0
∑F_{y}=0
∑torque = 0
3. The Attempt at a Solution
I don't have any forces working horizontally, and I don't think I need to set up an equation for the vertical forces because I don't have the mass of any of the beams. So that leaves me to set up a torque equation:
mgcos(22.6)(L/2) = kx(L/2)
m = (kx)/(gcos(22.6))
I need to solve for x, and when I originally attempted this problem I solved for x like this:
x = (L/2)tan(22.6)
But I didn't get the correct answer for m with this x, and I think it is because I solved for x as if it was perpendicular to the dashed line in the diagram. I am currently stuck trying to find x(change in length of spring from rest)
Attachments

7.6 KB Views: 426