Question

A cable is attached to a 1.50 kg block A, is looped over a fixed peg at C, and is attached to plate B.(Figure 1) The coefficient of static friction between the plate and the block is a = 0.240, the coefficient of static friction between the plate and the inclined plane is uB = 0.360, and the coefficient of static friction between the cable and the peg is uc = 0.290. If the plane's angle is 30.0 degrees, what is the maximum mass that plate B can have such that it does not slide down the plane?

When drawing the FBD of A, how to tell the direction of the friction force? also, when calculating T1 and T2(Using the equation T=Te^μβ)How to tell which force in the question is bigger?

Answer 1

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tes you want to at any downward o want to know how to decided the direction o friction of block A so assume their is no friction any surface and block & stall start making nevard , then block A will move upward assume fruction on ect on boily and definition pricction "Erinition force act opposite to the motion a A is molling upward then friction will to and block A is molling oct down word now, you want to know how to devle T and to For Plock & is melt is try to male downward and block A is try to molle upward now it hally > tno this direction motion T V to ☆ is try So (T2 trappen . now start of F.B.O of A ouestion NA > Ti UNA शिश

& Ey=0 1.5x9.81 cas 30 = NA E Fx=0 1 UNA - 1.5g sino = 0 NA= 12.74 3N T = .240x 12.743 +1.5x9-81X sin 30 T = 10.4158N F.B.O of Pully = 29 B = roeliau T2 = T UB P = 29 XT 10.4158 e = 25.936 N. F.B.O of Block B NB UANA TOMO NA но E Fg = 0 No= mBg as 30 +12.743 i mog sin 30 - Tz - UANA-UO NOT WB = 12.7113 +80 495 mo I M 3 x 4-81X sin 30° - 25.9036-28 X 12.743 - 036 (12.743 +8.495.00) mp = 4.213 ka
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