Question 12.SP.3: The two blocks shown start from rest. The horizontal plane a......

STRATEGY: You are interested in finding the tension in the rope and the acceleration of the two blocks, so use Newton’s second law. The two blocks are connected by a cable, indicating that you need to relate their accelerations using the techniques discussed in Chap. 11 for objects with dependent motion.

MODELING: Treat both blocks as particles and assume that the pulley is massless and frictionless. Because there are two masses, you need two systems: block A by itself and block B by itself. The free-body and kinetic diagrams for these objects are shown in Figs. 1 and 2, To help determine the forces acting on block B, you can also isolate the massless pulley C as a system (Fig. 3).

ANALYSIS: You can start with either kinetics or kinematics. The key is to make sure you keep track of your equations and unknowns.

Kinetics. Apply Newton’s second law successively to block A, block B, and pulley C.

Block A. Denote the tension in cord ACD by T_{1} (Fig. 1). Then you have

\underrightarrow{+}\Sigma F_{x}=m_{A}a_{A}\colon\qquad\qquad T_{1}=100a_{A}               (1)

Block B. Observe that the weight of block B is

W_{B}=m_{B}g=(300\ \mathrm{kg})(9.81\ \mathrm{m/s^{2}})=2940\ \mathrm{N}

Denote the tension in cord BC by T_{2} (Fig. 2). Then

+↓\Sigma F_{y}=m_{B}a_{B}\colon\qquad2940-T_{2}=300a_{B}\qquad (2)

Pulley C. Assuming m_{C} is zero, you have (Fig. 3)

+↓\Sigma F_{y}=m_{c}a_{c}=0\colon\qquad T_{2}- 2T_{1} = 0               (3)

At this point, you have three equations, (1), (2), and (3), and four unknowns, T_{1},~T_{2},~a_{B},~and~a_{A}. Therefore, you need one more equation, which you can get from kinematics.

Kinematics. It is important to make sure that the directions you assumed for the kinetic diagrams are consistent with the kinematic analysis. Note that if block A moves through a distance x_{A} to the right, block B moves down through a distance

x_{B}={\frac{1}{2}}x_{A}

Differentiating twice with respect to t, you have

a_{B}={\frac{1}{2}}a_{A}               (4)

You now have four equations and four unknowns, so you can solve this problem. You can do this using a computer, a calculator, or by hand. To solve these equations by hand, you can substitute for a_{B} from Eq. (4) into Eq. (2).

2940-T_{2}=300({\textstyle{\frac{1}{2}}}a_{A})

           T_{2}=2940-150a_{A}               (5)

Now substitute for T_{1}~and~T_{2} from Eqs. (1) and (5), respectively, into Eq. (3).

2940-150a_{A}-2(100a_{A})=0

2940-350a_{A}=0 \qquad \qquad a_{A}=8.40~\mathrm{m/s}^{2}  ◂

Then substitute the value obtained for a_{A} into Eqs. (4) and (1).

a_{B}={{\frac{1}{2}}}a_{A}={{\frac{1}{2}}}(8.40~\mathrm{m/s^{2}})\quad\quad\quad\quad\quad\quad\quad a_{B}=4.20~\mathrm{m/s^{2}}  ◂

T_{1}=100a_{A}=(100\ {\mathrm{kg}})(8.40\ {\mathrm{m/s}}^{2})\qquad\qquad T_{1}=840\ {\mathrm{N}}  ◂

Recalling Eq. (3), you have

T_{2}=2T_{1}\qquad T_{2}=2(840\ \mathrm{N})\qquad\qquad T_{2}=1680\,\mathrm{N}  ◂

REFLECT and THINK: Note that the value obtained for T_{2} is not equal to the weight of block B. Rather than choosing B and the pulley as separate systems, you could have chosen the system to be B and the pulley. In this case, T_{2} would have been an internal force.