Physics Torque Question?

Question

A lunch tray is being held in one hand, as the drawing illustrates. The mass of the tray itself is 0.200 kg, and its center of gravity is located at its geometrical center. On the tray is a 1.00 kg plate of food and a 0.275 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.
T = N (downward)?
F = N (upward) ? link to image:http://img251.imageshack.us/my.php?image=0968lr6.gif

Answer 1

Take a moment about the edge of the tray held by the hand
M=summation( RixFi) =0 where

Ri - lever of application of force Fi
Fi - forces in question

M= tray's + plate's +cup's + F's +T's=0
M= 0.2(.200)9.81 + 0.24(1.0)9.81 + 0.380 x 0.275 x 9.81 - 0.1F + 0.06T=0

Also the summation of all forces must be 0
Ft= tray's + plate's +cup's + F's +T's=0
(.200)9.81 + (1.0)9.81 + 0.275 x9.81 - F + T=0

Now we have two equations and two unknowns.
3.77 - 0.1F + 0.06T=0
14.47 - F + T=0
since
F= 14.47 +T
3.77 - 0.1(14.47 +T) + 0.06T=0
3.77 - 1.45 - .04T=0
T= (3.77 - 1.45 )/0.04 = 58N

F=14.47 +T=14.47 + 58 = 72.5N

Answer 2

The mass of the tray =m1=0.200 kg,

center of gravity of tray is located at its geometrical center

weight of tray W1=m1g= 0.1 g N.

distance of W1 from edge = r1 = 0.2 m

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Torque due to W1=T1 =W1r1=0.1g*0.2=0.02 g Nm (clockwise)
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weight of plate= W2=m2g=1.00g= g N

distance of W2 from edge= r2 =0.24 m
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Torque due to W2=T2 =W2r2 =g*0.24=0.24 g Nm (clockwise)

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weight of cup =W3=m3g=0.275 g N

distance of W3 from edge =r3=0.38 m
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Torque due to W3=T3 =W3r3=0.275g*0.38= 0.1045 g Nm (clockwise)

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force F exerted by the four fingers is upwards,at distance r= 0.1 m from edge
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Torque due to F=Tf =rF=0.1F Nm (anti clockwise)
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the force T exerted by the thumb is downwards at r!= 0.06 m from edge

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torque exerted by force T =Tt =r!T=0.06T ( clockwise)
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As the system is in equlibrium

upward force=sum of downward forces

F = T + W1 + W2 +W3

F = T + 0.2g + g + 0.275 g

F = T + 1.475 g_______________(a)
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anti clockwise torque = sum of clockwise torque

0.1 F=0.06T + 0.02 g + 0.24 g + 0.1045 g

F=0.6T + 0.2 g + 2.4 g + 1.045 g

F=0.6T + 3.645 g____________()b

from equation (a) and(b)

F =T + 1.475 g = 0.6T +3.645 g

0.4T=2.170 g

T = 5.425 g =5.425*9.8= 53.165 N

T = 53.165 N downwards
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F=T+1.475g = 53.165 +1.475*9.8=53.165+14.455 = 67.62 N

F= 67.62 N upwards
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