Class Managament I Halp HW 9 Begin Date: 10/9/2017 11:00:00 AM-Due Date: 10/23/2
ID: 1784551 • Letter: C
Question
Class Managament I Halp HW 9 Begin Date: 10/9/2017 11:00:00 AM-Due Date: 10/23/2017 11:00:00 AM End Date: 10/23 2017 11:00:00 AM (10%) Problem 9: You have been hired to design a family-friendly see-saw. Your design will feature a uniform board (mass M length L) that can be moved so that the pivot is a distance d from the center of the board. This will allow riders to achiere static equilibrium even if they are of different mass, as most people are. You have decided that each rider will be positioned so that his/her center of mass will be a distance xofte from the end of the board when seated as shown. You have selected a child of mass m (shown on the right), and an adult of mass n times the mass of the child (shown on the left) to test out your prototype. Etheexpertta.com 20% Part (a) Derve an expression for the torque applied by the adult rider (on the left m terms of gren quantities and vanables available in the palette. Assume counterclockwise is positive. Grade Summary Deductions 0% Potential 100% 4 5 6 Attempts remaining:- (5% per attempt) detailed view Xoffset N Submit Hint-1 I give up Hints. Ili deduction per hint. Hints remaining:- Feedback 1o deduction per feedbachk Submission History Hints Feedback Totals 06 00% 046 10% available in the palette. Assume counterclockwise is positive. - 20% Part (d) Detem me the distance d in terms of n g, and the masses and lenghs in the problem. variables available in the palette. Totals 20% Part b) Denve an expression for the torque applied by the child rider (on the right in terms of gren quantities and variables 20% Part (e) Denve an expression for the torque applied by the board in terms of gren quantities and arable a ailable in the palette. 20% Part (e) Determine the magnitude ofthe force exerted on the prot pom by the see-saw while m use m terms ofgren quantities andExplanation / Answer
a) torque = r x F
r = L/2 - d - xoffset
F = nmg
torque by adult = (0.5L - d - Xoffset)nmg
b) for child, r = 0.5L + d + xoffset
F = mg
torque = - (0.5L + d + xoffset)mg
c) torqur by board = - dMg
d) net torque will be zero.
(0.5L - d - Xoffset)nmg - (0.5L + d + xoffset)mg - dMg = 0
d(nm + m + M) = 0.5Lnm - Xofsetnm - 0.5Lm - xoffsetm
d = (0.5Lnm - Xofsetnm - 0.5Lm - xoffsetm ) / (nm + m + M)
e) F = mg + nmg + Mg