Couple quick questions

  • Thread starter squib
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  • #1
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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.
 

Answers and Replies

  • #2
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
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
Did you get the answer to #6? (the one with the rods and the two masses?)
 
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  • #5
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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
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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
haha I get it now, thanks
 
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  • #8
Did you get the "What angle does the vector angular momentum make with the axle?" question (#3 in capa)? Wouldnt it be zero?
 
  • #9
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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
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.
 
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  • #11
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any luck on that other one?
 
  • #12
I just changed my previous post
 
  • #13
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
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so they push up the same since same r?
 

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