Answer to Question #239507 in Physics for rshiii

Question #239507

27. Block 𝐴, 𝐵, and 𝐶 are placed as in figure and connected by ropes of negligible mass. Both 𝐴 and 𝐵 weigh 25 N each, and the coefficient of kinetic friction between each block and the surface is 0.35. Block 𝐶 descends with constant velocity (Figure 53). a) Draw two separate free-body diagrams showing the forces acting on 𝐴 and on 𝐵. b) Find the tension in the rope connecting blocks 𝐴 and 𝐵. c) What is the weight of block 𝐶? d) If the rope connecting 𝐴 and 𝐵 were cut, what would be the acceleration of 𝐶? Answer: b) 8.75Ñ, c) 30.8Ñ, d) 1.54m/s2


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Expert's answer
2021-10-04T10:07:50-0400


b) Write the equations by Newton's second law for all three bodies:

A:TAB=μmg=8.75 N.B:TAB+TBCμmgcos37°mgsin37°=0.C:Mg+TBC=0.A:T_{AB}=\mu mg=8.75\text{ N}.\\ B: -T_{AB}+T_{BC}-\mu mg\cos37°-mg\sin37°=0.\\ C:-Mg+T_{BC}=0.


From equation for B find TBC:


TBC=mg(μcos73°+sin37°+μ)=30.8 N.T_{BC}=mg(\mu \cos73°+\sin37°+\mu )=30.8\text{ N}.

c) From equation for C find Mg:


Mg=TBC=30.8 N.Mg=T_{BC}=30.8\text{ N}.

d) The acceleration is

C:Ma=Mg+TBC,B:TBCμmgcos37°mgsin37°=ma. a=1.54 m/s2,TBC=26.0 N.C:-Ma=-Mg+T_{BC},\\ B:T_{BC}-\mu mg\cos37°-mg\sin37°=ma.\\\space\\ a=1.54\text{ m/s}^2,\\ T_{BC}=26.0\text{ N}.


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