Question

In Figure (a), a uniform 44.4 kg beam is centered over two rollers. Vertical lines across the beam mark off equal lengths. Two of the lines are centered over the rollers; a 9.31 kg package of tamale is centered over roller B. What are the magnitudes of the forces on the beam from (a) roller A and (b) roller B? The beam is then rolled to the left until the right-hand end is centered over roller B (Figure (b)). What now are the magnitudes of the forces on the beam from (C) roller A and (d) roller B? Next, the beam is rolled to the right. Assume that it has a length of 0.833 m. (e) What horizontal distance between the package and roller B puts the beam on the verge of losing contact with roller A? (a) Number Units Units (b) Number Units (c) Number (d) Number Units (e) Number Units

Answer 1

44.489.8N 9.3189.8N RATRB = (44.4+9.31) g. 8N 22 RA+RB = 53.71x9.8N RA+RB = 526.358 N taking torque about a 44.489.8 +9:3/49:&xf = Rs.tk 108.78 + 45:619 = RR RB = 308.798 N / TRA = 217:56 N/ 44.4x9.8N 9.3189.8W 28 farque about A 44.4x9.880 + RATO + 9.31x9.0 lor 22.8095 = RB 7 RB 45.619 M RA= 526.358-45.619 RA - 480 739 NI distance moved by block from B = &- 0.833m. - 0.20825m 7 9 217.56N 308.798 N 480.739N 45.619 N 0.20825 m 7 9

Thanks for asking.

I hope it helps you.