Torque Questions

  • #1
hazybluesky
2
0

Homework Statement



Barry carries his tray of food to his favorite cafeteria table for lunch. The 0.5-m-long tray has a mass of 0.2 kg and holds a 0.4-kg plate of food 0.2 m from the right edge. Barry holds the tray by the left edge with one hand, using his thumb as the fulcrum, and pushes up 0.1 m from the fulcrum with his finger tips. How much upward force must his finger tips exert to keep the tray level?:bugeye:

g= 10 m/s^2

Homework Equations



Torque= Perpendicular Force x Lever Arm (T=Fxd)

Fw=mg

The Attempt at a Solution



Torque of tray...
Fw=mg
Fw= 0.2 * 10
Fw=2 N
d= length of tray? or 0.5 m? I'm not sure
T=Fxd
T=2*0.5
T= 1 <<<<<Torque of tray

Torque of plate...
Fw=mg
Fw= 0.4 * 10
Fw= 4 N
d= I have absolutely no idea!
T= 4 * ?? <<< Torque of plate

Then would you add the torque of the plate and tray together?
And then find the force using the that total torque? But what d value would you use?

Please, help me :frown:
Thank you!
-Ashley
 

Answers and Replies

  • #2
denverdoc
963
0
There are three torques to consider, two that are tending to rotate the tray out of his hands, and the third which must equal and oppose the sum of the first two. The torque of the tray can be considered a the mass X the distance to the center of the tray. See if that helps.
 
  • #3
hazybluesky
2
0
Is it the mass times the distance to the center of the tray? or the mass times gravity (to get a Force) times the distance to the center of the tray?
Thank you!

Also, can torques be added together? I think that the the two torques tending to rotate are that of the plate and tray, and that when added together, that is the torque needed to oppose the rotation. Is that correct?
 
Last edited:
  • #4
denverdoc
963
0
it the mass times gravity (force) X to the center of the tray, my oversight.
Yes, just ned to pick the same pivot point from which to reference the torques.
 

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