Question

3. A large board L = 6.00 m and m = 25.0 kg is laying so that d = 1.00 m of the board hangs off the roof of a building. A contractor puts his toolbox on the end of the board as shown. What is the greatest mass the toolbox can have and not cause the board to flip up and slide off the roof? A. 12.5 kg B. 25.0 kg C. 50.0 kg D. 245.0 kg

Answer 1

VW = mig - MG Given Information » large board of length (L) = 6.0m 2) Mass of board (M) = 25 kg 8) board is hanging at the roof at distance (d) = 1.00m Solution Let us consider mass of tool box be mi We consider that given rod is of uniform density thus centre of mass (COM) is alting at the middle of rod at 3 m.]

Tool box to not flip up and slide off the mof. torque arting at com & at other end must be equal and opposite. Normal force IN - Mg - Weight of rod) acting upwords LLI Torque due to Normal forte acting upwords = torque weight of tool box acting down words due to Normal forre = N = Mg - - Weight of tool box = W = mig h- LLLL LLLLLLLLLLL :. Torque (c) a distance from a forced aris of rotation ? 7 XF -rf sino Herer, f are general terms of distance & force respectively NX (-d) = wxd nende N (1 d) = wd na sino = singo = 1 substing value of 0 & ② in above equation. Mg (3-1) = m, d (1) 2 mi= 20 = 2 x 25 = 50 kg Thus masstemi) = 50 kg so that tool box will not flip up glide off the roof. : -

4 liven Information. m = 0.650 kg R = 0.1tom Vo = 9.650 m/s t = 4.110s. As shown in above diagram, consider a ball rolling without slipping.. Since the ball is moving with constant velocity to distance covered in time t will be d = Vot = 3.650 84.110 Distance moved by centre of the ball is d which is same as above, It can be given by . : d=RO where o is total angle covered to in Rotation Ris Radius of ball 1. d = 3.650 X4:11 - RO 3.650 X4-11 - 0 R 0 - 3.650X 4-11 - 0.70 To No. of Rotation will be o = 3.650x4.1) - 27 0.170 X2 T of Rotations ball made = 13.70

Given Information m = 1230 kg R = 0.230m initial velocity u = 0 m/s angular acceleration (x) = g.11 rad/5² time (t) = 5.345. L from Kinematical equation of motion we know that v=utat a an..0 Wheel's are rotating thus relation between lineay and angular acceleration is given by - a = RX . . - - - substituting a in equation o v=ut Rat but uzo V=Rat = 0.230 x g. 11 X 5.34 - 11.19 m/s ~11.2 m/s & Velocity of car after time t is 11.2 m/s) Answer option Bl