Lunch tray and torque

1. gamesandmore

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 1-kg plate of food and a 0.350-kg 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 Counter-clockwise 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. gamesandmore

32
Anyone??? I need this as soon as I can.

3. PhanthomJay

6,151
You have somewhat of an idea of the correct approach, but when you sum torques about the left end, you have to include the torque provided by the thumb as well as the fingers. It is better to sum torques about the thumb, because then that force will not enter into the torque equation because it has no torque. Then apply Newton 1 in the y direction to determine the thumb force.

4. gamesandmore

32
im confused... sum up at the left end?

Last edited: Nov 28, 2006
5. PhanthomJay

6,151
Newton's first law requires that the torques must sum to 0 (Clockwise torques = counterclockwise torques); AND, the forces in the vertical direction must sum to 0 (Forces up = forces down). Now try summing torques about the thumb and see what you get for Ff. You'll have to do a little math to get the proper distances from the load points to the thumb, but so what?

Last edited: Nov 28, 2006
6. gamesandmore

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. PhanthomJay

6,151
You had the right approach, stop stressing. Look at the clockwise torques about the thumb. I'll start you off withthe cup of coffee. Its torque about the thumb is

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. gamesandmore

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?

Last edited: Nov 28, 2006
9. PhanthomJay

6,151
I didn't check your math, but once you get F, then apply Newton 1: weight of tray plus weight of cup plus weight of plate plus the thumb force T = F. Clock is ticking, I usually don't do all the work.....

10. PhanthomJay

6,151
Gee, now you got me stressed. How we doing for time? Check your math. .35(9.8)(.33) + 1(9.80)(.19) + (.18)(9.8)(.150) = F(.050). Solve for F. , then solve for T.

11. gamesandmore

32
Yes, I'm sorry I didn't reply again.
Thank You very much, I got full credit for the problem. :)

12. Haibane

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

-Haibane

Last edited: Dec 6, 2006

6,151