Question

1768 Mg Ns 36° 36 L cos36 Two friends are carrying a crate (with uniform density) up a flight of stairs. The crate is being carried parallel to the slope of the stairs which make a 36 degree angle with the floor NOTES IMAGES DISCUSS UNITS STATS HELP Part Description Answer Chk History Assume static equilibrium Determine the moment of the weight of the crate about the lower suppot point. (CCW is positive) 1. - Mg (0.625 m) 2. Mg (0.625 m cos 36) 3. Mg (0.625m cos 36°) +Mg(0.25m sin 36°) 4. - Mg (0.625m sin 36°) +Mg (0.25m cos 36°) 5. Mg (0.625m cos 36°) +Mg(0.25m tan 36) A. 6.67 pts. 100%| # tries: 0 Show Details Hints: 0,0 If the force each person applies is vertical, is it less weight for 1. the person above on the 2. the person below on the stairs 6.67 pts, 100% , tries: 0 Show Details stairs 3. Does not matter

Answer 1

Torque of weight 'mg' about p is By resolving mg we get tau = mg sin theta times 0.25 - mg cos theta = 1.25/2 tau = 2000 [sin times 0.25 - cos 36 times 0.625] tau = -717.4 N-m -mg cos 36 times 0.625 + mg (0.25 sin 36) B) Torque about 0 By N_A & N_B is equal & opposite so N_A = N_B so does not matter