Question

(25%) Problem 4: You have been hired to design a family-friendly see-saw. Your design will feature a uniform board (mass M = 15 kg, length L = 11 m) that can be moved so that the pivot is a distance d from the center of the board. This will allow riders to achieve 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 xoffset = 19 cm from the end of the board when seated as shown. You have selected a child of mass m = 26 kg (shown on the right), and an adult of mass n =5 times the mass of the child (shown on the left) to test out your prototype.

offset offset

Part (a) Determine the distance d in m. $sig.gif?tid=6C65-1D-58-44-90B8-19151$

50% Part (b) Determine the magnitude of the force exerted on the pivot point by the see-saw in N.

Answer 1

solution Part (o) e For the equilibrium 26x9.8x[ 4 -0.19+d7+1569.8 xd. = 5X26X9,8x/ 1-0019-d] OR 254.8*(5,31+q) +1472 = 1274 x( 563)- → 1352.988 + 254,88 +1470 = 6764,94-12744 → 167508d = 5411,952 = d = 3.23 m a = 3,23 7 m Part 2 > Magnitude af force by the piot = 4. M = 126 + 130 +15)*9.8 = 171 *9.8 = 167578 N. 167568 IN