Calculating Forces on a Lunch Tray

[gamesandmore]

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?

 

[gamesandmore]

Anyone? I need this as soon as I can.

 

[PhanthomJay]

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?

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.

 

[gamesandmore]

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

 

[PhanthomJay]

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

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?

 

[gamesandmore]

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.

 

[PhanthomJay]

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.

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.

 

[gamesandmore]

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?

 

[PhanthomJay]

alright, I got F: 79.0 N
Thank you!

T I am confused on right now... would it be the same value as F?

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...

 

[PhanthomJay]

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?

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.

 

[gamesandmore]

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

 

[Haibane]

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 I've 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

 

[PhanthomJay]

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 I've 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

Check your givens and math:
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?

 

[Haibane]

Well I am 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 I am 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.