# Couple quick questions

1. Apr 15, 2005

### squib

1. Two builders carry a sheet of drywall up a ramp. Assume that W = 1.70 m, L = 3.20 m, = 18.0°, and that the lead builder carries a weight of 137.0 N (30.8 lb). What is the weight carried by the builder at the rear ?

I know torque and all forces add up to zero, as the drywall is not rotating, but don't I need to know the mass of the drywall??

2. A rotating uniform cylindrical platform of mass 284 kg and radius 5.45 m slows down from 3.81 revolutions per second to rest in 18.2 s when the driving motor is disconnected. Calculate the power output of the motor required to maintain a steady speed of 3.81 revolutions per second.

I tried Ke/time, but that didn't work... so out of ideas.

2. Apr 15, 2005

### ninjagowoowoo

haha I have those problems too. For the second one:

1. Find alpha using the w=wnot + alphaT And find I using (1/2)MR^2
2. You now have enough information to find the torque due to friction (I times alpha)
3. So to keep the disk spinning at a constant omega, the net torque must equal the torque due to friction.
4. Then use the formula for angular Power (torque times omega)

I still havent found the answer to the first question.... hmmm

3. Apr 15, 2005

### ninjagowoowoo

I believe that for the first one , you need to find the torque exerted by each of the builders. Then set the torqus equal to each other. However I don't know where to get the torque from (center of rotation). But once you do, you can use torque = rFsin(theta). So basically all you need to do is find the radius and the force. Anyway, I still havent quite got the answer, but i'm getting closer.

4. Apr 15, 2005

### ninjagowoowoo

Did you get the answer to #6? (the one with the rods and the two masses?)

Last edited: Apr 15, 2005
5. Apr 15, 2005

### squib

Yeah... umm, did that one yesterday, so let me see if i remember...

The total force downward is 1.5mg.
Then, find the center of mass for each block, and calculate the torque that is exerted onto each side, using the other side as the pivot (ie, use the left support as the pivot for calculating the force on the right side).

Then find the ratio of torques. My total came out to 12/8, which is 1.5, as from the 1.5mg in the beginning. L does not need to be known because it is divided out. That make any sense? I'm afraid it probably didn't.

6. Apr 15, 2005

### squib

I came out with right: (1/4)l(.5mg) + (1/2)lmg = (5/8)mgl
Left: (3/4)l(.5mg) + (1/2)lmg = (7/8)mgl

7. Apr 15, 2005

### ninjagowoowoo

haha I get it now, thanks

Last edited: Apr 15, 2005
8. Apr 15, 2005

### ninjagowoowoo

Did you get the "What angle does the vector angular momentum make with the axle?" question (#3 in capa)? Wouldnt it be zero?

9. Apr 15, 2005

### squib

think of it as the angle that goes between the masses, through the centerpoint of the rod, like:

*----|
|
O
|
|----*

the angle through the 0(center of mass)

10. Apr 15, 2005

### ninjagowoowoo

sweet thanks a bunch. This builder problem is frustrating. So far I'm assuming the center of rotation is at the center of mass. So then all i need are the angles between the radius and the forces from the builders. This could be used in Fsin(theta) = F_2sin(theta_2). Any ideas on how to get the angles? I don't remember enough geometry... But does what I'm doing sound right?

Oh and the R's cancel out bc the "radiuses" from the COM are equal.

Last edited: Apr 15, 2005
11. Apr 15, 2005

### squib

any luck on that other one?

12. Apr 15, 2005

### ninjagowoowoo

I just changed my previous post

13. Apr 15, 2005

### ninjagowoowoo

did you get the very last problem? vertical component of the force exerted by the hinge on the beam? What did you use to calculate that...?

14. Apr 15, 2005

### squib

so they push up the same since same r?