Answer to Question #170277 in Physics for RHYLYN

Question #170277


Blocks A, B and C are placed as shown in figure and connected by ropes of negligible mass. Both A and B weigh 25.0 N each, and the coefficient of kinetic friction between Block A and B and the surface is 0.35. Block C descends with constant velocity. Assume massless and frictionless pulleys.


a. Draw the free-body diagrams of each of the objects.


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?




1
Expert's answer
2021-03-16T08:33:12-0400

a. Draw the diagram:




b. Consider tension between blocks C and B. Blocks A and B can be treated as a single body since they are connected by a string that does not stretch. According to Newton's second law:


Ox:TBCfAfB=(MA+MB)a,               TBCμNAμNB=(MA+MB)a.Oy:NAMAg=0,Oy:NBMBg=0.Ox:T_{BC}-f_A-f_B=(M_A+M_B)a,\\ \space\space\space\space\space\space\space\space\space\space\space\space\space\space\space T_{BC}-\mu N_A-\mu N_B=(M_A+M_B)a.\\ Oy: N_A-M_Ag=0,\\ Oy: N_B-M_Bg=0.

Therefore:


TBCμMAgμMBg=(MA+MB)a.TBC=μ(gMA+MBg)=17.5 N.T_{BC}-\mu M_Ag-\mu M_Bg=(M_A+M_B)a.\\ T_{BC}=\mu(g M_A+M_Bg)=17.5\text{ N}.

Since block A moves at a constant speed as well, the tension between A and B is equal to the force of friction of A:


TAB=fA=μNA=μMAg=8.75 N.T_{AB}=f_A=\mu N_A=\mu M_Ag=8.75\text{ N}.



c. Since block C descends at constant speed (zero acceleration), no other forces except tension and gravity act on it:


Oy:TBCgMC=MCa=0,MC=TBCg=1.786 kg.Oy:T_{BC}-gM_C=-M_Ca=0,\\ M_C=\frac{T_{BC}}{g}=1.786\text{ kg}.

d. With block B only according to Newton's second law:


Ox:TBCfB=MBa,Oy:TBCgMC=MCa.Ox: T_{BC}-f_B=M_Ba,\\ Oy:T_{BC}-gM_C=-M_Ca.

This is a system with two unknowns: TBCT_{BC} and aa. The solution for acceleration is


a=gMCμgMBMB+MC=2.018 m/s2.a=\frac{gM_C-\mu gM_B}{M_B+M_C}=2.018\text{ m/s}^2.

Need a fast expert's response?

Submit order

and get a quick answer at the best price

for any assignment or question with DETAILED EXPLANATIONS!

Comments

No comments. Be the first!

Leave a comment