Constrained Motion of 7 Masses and 3 Pulleys

[mingyz0403]

Homework Statement

Block B starts from rest, block A moves with a constant acceleration, and slider block C moves to the right with a constant acceleration of 75 mm/s2. Knowing that at t = 2 s the velocities of B and C are 480 mm/s downward and 280 mm/s to the right, respectively, use kinematics to determine:
a. The acceleration of A and B
b. The initial velocities of A and C
c. The change in positions of slider C after 3 s.

Relevant Equations

Motion equation.
Conservation of string

How do you set the direction for this problem? Do you look at the motion of the block? If you set right positive, does that automatically makes upward motion positive? I set right as positive and up as positive for this problem. However, my answer was wrong. Textbook solution set right as positive and down as positive. I don't understand why setting right as positive and up as positive wrong for this problem.

Many thanks for your help.

 

[Lnewqban]

... I don't understand why setting right as positive and up as positive wrong for this problem.

I don’t believe that setting the directions of positive one way or the other makes a difference in the result.
The sign of the value that you obtain at the end always tells you whether or not your initial assumption regarding the direction of that magnitude was correct.

Your error was assigning a negative value to $4a_b$.
When in doubt, it is better to assume all as positive and let the result decide.

The original data is telling us that:
1) The length Xc is growing as block C is moving to the right.
2) The four lengths Xb are growing as block B is moving downwards.
3) The three lengths Xa are shrinking as block A can only move upwards as result of conditions 1) and 2) above.
$$\begin{align}
a_C+4a_B+3a_A&=0\nonumber\\
75+4(480/2)+3a_A&=0\nonumber\\
-75-960&=3a_A\nonumber\\
-1035/3&=a_A\nonumber\\
a_A&=-345~mm/s^2\nonumber\\
\end{align}$$
Note that the negative sign here means that the actual direction of the acceleration of block A is opposed to the originally assumed.