i posted this once but i did what i was told nad still did not get right answer can someone please help?!?!?!!

Question

ok this is alot of numbers and information so here go
a lunch tray is being held by one hand. teh mass of the tray is .220 kg... its center gravity is at its center. on the tray is a 1 kg plate and a a .335 kg cup of coffee. obtain force t (exerted down ) by thumb and force f (exerted up ) by four fingers. both forces act perpendic. to tray, which is held parrallel to ground.. ...
other info....
length of tray is .400m
length from beginning of tray to .......
cup is .380m
plate is .240 m
four fingers is .100m
to thumb is .0600m

ok i tried settin up net force = 0 and net torque =0 then solving for the different components and things... someone i keep getting the wrong answer though... i dont know if i am messing up the formulas or if i am just missing something?!?!?! i know torque is force time lever arm so i dont know where i am going wrong could someone walk me through this so maybe i can see my mistake it would be great help it could just be a small error thank you

Answer 1

First, draw a diagram. In this case, a line on the X-axis to represent the tray will do. On this line, there will be four forces pulling it down and one force holding it up. The one holding is up is from the four fingers on the bottom. The easiest way to solve this is to set X=0 where the fingers are.

Now figure out magnitudes of the known forces:

W_tray=0.22kg*9.8m/s^2
W_cup=0.335kg*9.8m/s^2
W_plate=1.0kg*9.8m/s^2

So on your line (with X=0 set to where the fingers are), you will have a force W_tray acting down at X=0.1m (0.2m at the center of the tray minus 0.1m of the fingers from the left edge due to redefining the origin). You have W_cup acting down at X=0.280m (0.38m-0.10m). You have W_plate acting down at X=0.14m (0.24m-0.10). These 3 forces at various points to the right of the finger will generate the total clockwise moment.

Now the only moment to counter this is the moment from the thumb. The thumb will apply an unknown force F_thumb at 0.04m to the left X=0. So when you write out the moment balance equation, you end up with this:

0.04*F_thumb=0.1*W_tray+0.28*W...

You can solve for F_thumb, which is also acting downward. The moments are balanced so the plate doesn't rotate.

Now that you have all the forces acting downward, you can find the force acting upward from the fingers on the bottom of the tray. You just sum up all the forces acting downward, and the opposing force is just the same magnitude in the opposite direction. Now the forces are also balanced such that it doesn't translate.

Note that you don't have to set X=0 where the fingers are. You can set origin at any arbitrary location. Your moment equations will change accordingly with different lever arms. The reason that I did it this way is that it makes the moment due to the force of the fingers acting upwards disappear in your moment equation, allowing you to solve for one of the unknowns directly. You can set the origin where the thumb is and get similar simplification. If you set the origin to the edge of the tray or somewhere else, you have the added complexity that you have to solve the two equations simultaneously for the two unknowns.