Calculating Forces on a Lunch Tray

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In summary: F = T + 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.Yes, I'm sorry I didn't reply again.Thank You very much, I got full credit for the problem. :)In summary, the forces acting on the tray are the thumb force of 40.1 Newtons, the four finger force of 18
  • #1
gamesandmore
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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.

09_68.gif


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?
 
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  • #2
Anyone? I need this as soon as I can.
 
  • #3
gamesandmore said:
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.

09_68.gif


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.
 
  • #4
im confused... sum up at the left end?
 
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  • #5
gamesandmore said:
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?
 
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  • #6
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
gamesandmore said:
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.
 
  • #8
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?
 
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  • #9
gamesandmore said:
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...
 
  • #10
gamesandmore said:
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.
 
  • #11
Yes, I'm sorry I didn't reply again.
Thank You very much, I got full credit for the problem. :)
 
  • #12
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 :bugeye:

Thanks ahead of time!
-Haibane
 
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  • #13
Haibane said:
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 :bugeye:

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?
 
  • #14
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.
 

1. What is a lunch tray and how does it relate to torque?

A lunch tray refers to a flat, rectangular tray commonly used to serve food in cafeterias and school lunchrooms. In physics, torque is a measure of the force that causes an object to rotate around an axis. The concept of torque can be applied to a lunch tray when it is being carried and tilted, as the weight of the food and the position of the tray's center of mass can affect the amount of torque exerted on the person carrying it.

2. How does the design of a lunch tray affect its torque?

The design of a lunch tray can affect its torque in several ways. For example, a lunch tray with a wider base and lower center of mass will be more stable and have less torque compared to a tray with a narrower base and higher center of mass. Additionally, the material of the tray can also impact its torque, as a heavier tray will have more torque compared to a lighter one.

3. Can torque be used to prevent spills on a lunch tray?

Yes, torque can be used to prevent spills on a lunch tray. By adjusting the position of the tray's center of mass, one can decrease the amount of torque exerted on the tray, making it less likely to tip over. Additionally, by evenly distributing the weight of the food on the tray, the overall torque can be reduced.

4. How does the type of food on a lunch tray affect its torque?

The type of food on a lunch tray can affect its torque in various ways. Generally, heavier and denser foods will increase the overall weight of the tray, resulting in more torque. Additionally, the placement of the food on the tray can also impact its torque, as food placed closer to the edge will increase the distance between the tray's center of mass and the axis of rotation, resulting in more torque.

5. How can the concept of torque be applied to other everyday objects?

The concept of torque can be applied to many everyday objects, such as doors, wrenches, and even bicycles. Any object that rotates around an axis or pivot point can be affected by torque. Understanding the principles of torque can help in the design and use of these objects to ensure stability and prevent accidents.

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