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Lunch tray and torque 
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#1
Nov2806, 08:08 PM

P: 32

A lunch tray is being held in one hand, as the drawing illustrates. The mass of the tray itself is 0.180 kg, and its center of gravity is located at its geometrical center. On the tray is a 1kg plate of food and a 0.350kg cup of coffee. Obtain the force T exerted by the thumb and the force F exerted by the four fingers. Both forces act perpendicular to the tray, which is being held parallel to the ground.
I don't even have a clue where to start?? Here is what I tried though, but I'm not sure: Sum of Counterclockwise Torques = Sum of Clockwise Torques Tf = Tplate + Tcup + Ttray then: Ff(.1m) = (1kg*9.80m/s^2)(.24m)+(.35kg*9.80m/s^2*.38m)+(.18kg*9.80m/s^2*.2m) and ended up with F being equal to: 40.1... but I'm not sure? 


#2
Nov2806, 08:46 PM

P: 32

Anyone??? I need this as soon as I can.



#3
Nov2806, 08:49 PM

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#4
Nov2806, 08:58 PM

P: 32

Lunch tray and torque
im confused... sum up at the left end?



#5
Nov2806, 09:11 PM

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#6
Nov2806, 09:21 PM

P: 32

I understand the part about newton's first law...
But I cannot figure out what you mean... This is due in 40 minutes and I'm stressing out about it. 


#7
Nov2806, 09:34 PM

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0.35(9.8)(.330) = 1.13 Now do the tray and plate torques about the thumb, and add 'em up with the coffee torque detrmined above, then set the result equal to the fingers counterclockwise torque of Ff(0.50), and solve for Ff. Remember, the distance to use in finding the torques is the distance form the load to the thumb. 


#8
Nov2806, 09:38 PM

P: 32

alright, I got F: 79.0 N
Thank you!!!! T I am confused on right now... would it be the same value as F? F = T + Fplate + Fcup + Ftray? edit: I tried that and got 64 as an answer, sound right? 


#9
Nov2806, 09:44 PM

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#10
Nov2806, 10:00 PM

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#11
Dec106, 07:15 PM

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Yes, I'm sorry I didn't reply again.
Thank You very much, I got full credit for the problem. :) 


#12
Dec606, 10:29 PM

P: 5

I apologize for bringing up an old thread but I have more or less the same problem with this equation, even after looking at it and trying to solve it.
It is the same except that the mass of the tray 0.200 kg, and the coffee is 0.250 kg. Now Ive tried working it out, factoring in the fact that the cup of thumb is at a distance of .0600m which means using this as the axis Id have to remove .0600 for every lever arm which gives me. (F*.04) = ((9.8*1kg)*.04)+((9.8*.250)*.32)+((19.8*.180)*.34) which leads to (F*.04) = 1.764 + .784 + .59976 or just (.60) which = 3.148 and then is divided by .04 to = 78.7.... when the correct answer for F supposed to be 70.6. What am I missing or doing wrong? This problem is driving me up the wall Thanks ahead of time! Haibane 


#13
Dec706, 12:15 AM

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F(.04) = 9.8(1)(.18) + (9.8(.25)(.32) + 9.8(.2)(.14) Solve F = 70.6N You had some typos in there, but mostly you messed up that last term. You had the weight of the tray acting at the far right end, when actually, since the load is uniformly distributed, it acts at the c.g of the tray, which is dead center at .20m from the right end or left end, that is, .14m from the thumb, not .34m. Do you see why? 


#14
Dec706, 08:33 AM

P: 5

Well Im going out on a limb here in terms of my guess, but I get the feeling that because the cg of the tray is in the center, rather than the lever arm running all the way to the end of the try, it would run to the center at .20m. And consequently .20  .06 = .14? If so then Im glad I finally got it, haha.
And also thanks for correcting my typos, by the time I wrote this post last night I was brain dead as Id spent the past few hours bashing my head into the wall over said problem. 


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