Your crazy uncle Mike has odd preferences for where weights should hang on the f

Question

Your crazy uncle Mike has odd preferences for where weights should hang on the force table and an obssession with balance. He's only happy if a 130g weight hangs from 0 degrees, and a 190g weight hangs from 137 degrees. Where does the third weight need to hang in order for the central disk to be balanced? How much should it weigh?

a) Draw a vector diagram (to scale) of the two given forces on the disc, labeling the weights and angles.

b) Calculate the predicted weight and angle (ignore uncertainty) of the third force. Draw and label this third force to scale in the correct place on your vector diagram.

c) The percent uncertainty = 2%; use this to calculate the mass uncertainty for the two given masses and propogate this uncertainty to find the uncertainty in the predicted magnitude for the third force (ignore angle uncertainty). Is this predicted third force consistent with the force from the third mass (with uncertainty) that you actually found in b?

Explanation / Answer

F1y = -130
F1x = 0

F2y = 190*sin43 = -130
F2x = -190*cos43 = -139

Fy = F1y+F2y = -130-130) = -260
Fx = F1x+F2x = -139
F = (-139 ; -260)

-F = (139 ; 260)

260 g are needed

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