Question

Problem #1 m1 m2 Two blocks mı = 4 kg and m2 = 9 kg are initially arranged as shown in the figure. They are tied to a massless rope going around the pulley. The pulley has a form of a cylinder with a mass of M = 8 kg and radius of R = 40 cm. Both the incline and the horizontal surface have a coefficient of kinetic friction x = 0.15. The incline is at the angle = 55° with respect to horizontal. The blocks are released from rest and begin to accelerate such that ma is moving down the incline and mı moves horizontally. Find (a) the tension (in N) in the rope on the left side of the pulley; (b) the tension (in N) in the rope on the right side of the pulley; (c) the acceleration (in m/s) of the block mı; (d) the angular acceleration in rad/sec) of the pulley; (e) the speed (in m/s) of the block m2 after it moves for At = 2 sec.

Answer 1

DATE! solution Free Joo dy diagram as acceleration a ad РТ, To av mg fo zgsins- miguk (B) te meglasses mia = Ti-fi (0) - 0) R. mgesso from a mia = Ti-miguk From b) a = a a Id= T2 R-TIRI Ia R = (2-TR IQ- (T2-T,) R2 MR?az (T2-T1) R² ma= Tz-TI Frem (0) {Iam = MRZY mia- mza magsino-fz-Tz megsîno-mg casa ulk - Tz -(1)

Now we addling (12, (01 ) and (1) equoitions. get mia + ma+mza = magsing-miguk- meg casosluk mi+me+m) a = my megsinos -(mi + m2 caso) + m2 casoguk >)( 4+9+8) a = 9x10xsinss.-(4+9casso )x10x 0.15 a = 2.86 mis ca). 4x 2.86 mis - Ti - 4x10x0.15 From 131 TI = 17.44 N Tension in the left side of pulley is 17.44 N (b) 8x 2.86 = T2-17.44 & from (11) T2 = 40.32 N Tension in the right side of pulley is 40. 320 (e) acceleration of block m, is 2.86 mis2 d2 angular acceleration of the pulley isdag =2-86 400m - 7.15 rotts 0.40mi (e) {u=0, t=25/ so, V = 0 + 2.8 6 X 2 = 5.72 mis Speed of block mi is 5072 mis of=2086 v = ut at