Answer to Question #170277 in Physics for RHYLYN

Question

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?

Answer 1

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:

$$Ox:T_{BC}-f_A-f_B=(M_A+M_B)a,\\ \space\space\space\space\space\space\space\space\space\space\space\space\space\space\space T_{BC}-\mu N_A-\mu N_B=(M_A+M_B)a.\\ Oy: N_A-M_Ag=0,\\ Oy: N_B-M_Bg=0.$$

Therefore:

$$T_{BC}-\mu M_Ag-\mu M_Bg=(M_A+M_B)a.\\ T_{BC}=\mu(g M_A+M_Bg)=17.5\text{ N}.$$

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:

$$T_{AB}=f_A=\mu N_A=\mu M_Ag=8.75\text{ N}.$$

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

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

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

$$Ox: T_{BC}-f_B=M_Ba,\\ Oy:T_{BC}-gM_C=-M_Ca.$$

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

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