Question 12.SP.4: Collar A has a ramp that is welded to it and a force P = 5 l......

STRATEGY: The principle you need to use is Newton’s second law. Because a block is sliding down an incline and a cable is connecting A and B, you also need to use relative motion and dependent motion.

MODELING: Model A and B as particles and assume all surfaces are smooth. As usual, start by choosing a system and then drawing a free-body diagram and a kinetic diagram. This problem has two systems, and you need to be careful with how you define them. The easiest systems to use are (a) collar A with its pulley and the ramp welded to it (system 1) and (b) block B and the pulley attached to it (system 2), as shown in Fig. 1. The free-body and kinetic diagrams for system 1 are shown in Fig. 2. The free-body and kinetic diagrams for B are a little trickier, because you don’t know the direction of the acceleration of B.

Kinematics for Block B. Express the acceleration a_B of block B as the sum of the acceleration of A and the acceleration of B relative to A. Hence,

\mathbf{a}_{B}=\mathbf{a}_{A}+\mathbf{a}_{B/A}

Here \mathbf{a}_{B/A} is directed along the inclined surface of the wedge. Now you can draw the appropriate diagrams (Fig. 3). Note that you do not need to use the same x–y coordinate system for each mass, because these directions are simply used for obtaining the scalar equations.

ANALYSIS: You can obtain a scalar equation by applying Newton’s second law to each of these systems.

a. System 1:

\underrightarrow{+}{\Sigma}F_{x}=m_{A}a_{A_{x}}\qquad N_{A}-N_{B}\cos50^{\circ}+2T\cos40^{\circ}=0            (1)

b. Block B:

+ \searrow \Sigma F_{x}=m_{B}a_{B_{x}}\qquad-2T+W_{B}\sin40^{\circ}=m_{B}a_{B/A}+m_{B}a_{A}\sin40^{\circ}            (3)

+\nearrow \Sigma F_{y}=m_{B}a_{B_{y}}\qquad N_{B}-W_{B}\cos{40}^{\circ}=-m_{B}a_{A}\cos{40}^{\circ}            (4)

You now have four equations and five unknowns (T, N_{A},~N_{B},~a_{A},~and~a_{B/A}), so you need one more equation. The motions of A and B are related because they are connected by a cable.

Constraint Equations. Define position vectors as shown in Fig. 4. Note that the positive directions for the position vectors for A and B are defined from the kinetic diagrams in Figs. 2 and 3. Assuming the cable is inextensible, you can write the lengths in terms of the coordinates and then differentiate.

Constraint equation for the cable: x_{A}+2x_{B/A} = constant

Differentiating this twice gives

a_{A}=-2a_{B/A}            (5)

You now have five equations and five unknowns, so all that remains is to substitute the known values and solve for the unknowns. The results are N_{A} = −0.1281 lb, N_{B} = 0.869 lb, T = 0.281 lb, a_{A} = −13.46 ft/s², and a_{B/A} = 6.73 ft/s².

T = 0.281 lb ◂

REFLECT and THINK: In this problem, we focused on the problem formulation and assumed that you can solve the resulting equations either by hand or by using a calculator/computer. It is important to note that you are given the weights of A and B, so you need to calculate the masses in slugs using m = W/g. The solution required multiple systems and multiple concepts, including Newton’s second law, relative motion, and dependent motion. If friction occurred between B and the ramp, you would first need to determine whether or not the system would move under the applied force by assuming that it does not move and calculating the friction force. Then, you would compare this force to the maximum allowable force μ_{s}N.