Two baby pandas play with a toy that consists of a rope and an ideal pulley, as shown in the picture. If another baby panda jumps on one side of the rope:

(a) Calculate the acceleration of the pandas in terms of \(g\).

(b) Find the tension on the rope in terms of the mass of one baby panda and \(g\).

Assume that all the baby pandas have the same mass.

1. (a) The first part of the problem asks us to determine the  of the pandas in terms of the gravitational constant \(g\).

Question 1 of 10

2. To approach the question, and as with all the problems involving forces and acceleration, we'll start by making a free-body diagram for the pandas. Then we will use   to relate the forces to the acceleration of the pandas.

Question 2 of 10

3. Let's begin with the free-body diagram. We'll make a free body diagram for the panda on the left, and we'll treat the two pandas on the right as if they were a single system (when solving a problem, we can always group two or more objects in a single system, but if the two objects do not move in the exact same manner, then doing this might not be convenient).

We know that on the left panda and the right pandas there is a tension force directed upwards in both cases. The magnitude of the tension on both sides is , since we are considering that the pulley on the top is frictionless and the rope does not stretch. The other force we identify is the weight of each panda, which is directed and has magnitude \(mg\), where \(m\) is the mass of one panda and \(g\) is the gravitational acceleration on Earth.

Question 3 of 10

4. Hence, the force diagram for the left panda is shown in figure 1.

Question 4 of 10

5. We'll use Newton's second law on the Y axis for the panda of the left to write

 

Question 5 of 10

6. We'll use Newton's second law on the Y axis for the panda of the left to write

 

Question 6 of 10

7. where \(a_L\) is the acceleration of the left-hand side panda. Dropping the vector notation because all the quantities are along the same axis, we get from equation \eqref{newtonl}

\begin{equation}
\tag{3}
\label{newtonl222222}
T-mg=ma_L.
\end{equation}

We can do the same for the pandas on the right and write Newton's second law for the Y axis as

 

Question 7 of 10

8. where on the right hand side of the equation above we wrote the total mass of the pandas \(m+m\) and their acceleration, denoted by \(a_R\). Dropping the vector notation because all the quantities are along the same axis, we get, from equation \eqref{newtonr},

\begin{equation}
\tag{5}
\label{newtonr2}
T-2mg=2ma_R,
\end{equation}

where we have also summed the weights to get \(-2mg\).

So far, we have two equations and three unknowns:

Question 8 of 10

9. Thus, for us to be able to solve the problem, we need  involving these unknowns. This additional equation(s) will be a relation between the accelerations of the panda on the left with the panda on the right.

Notice that the pandas are hanging on the same (ideal) rope, which passes through a simple pulley. It is easy then to see that if the pandas on the right go down, the panda on the left must go and vice-versa. Furthermore, if the pandas on the right fall 2 meters, the panda on the left will go this same 2 meters. The same idea applies to the velocity and acceleration. Taking into account that the acceleration is a , we know that the magnitudes of the accelerations \(a_R\) and \(a_L\) must be  .

Question 9 of 10

10. This leads us to write the intuitive yet useful relation

\begin{equation}
\tag{6}
a_R\,\hat{\textbf{j}}=-a_L\,\hat{\textbf{j}}.
\end{equation}

Dropping the vector notation and focusing on the components, we get

\begin{equation}
\tag{7}
\label{ligadura}
a_R=-a_L.
\end{equation}

We can now use the relation given in equation \eqref{ligadura} together with equations \eqref{newtonl2} and \eqref{newtonr2} to solve for the acceleration. Let's start by solving for \(T\) in equation \eqref{newtonl2} to get

Question 10 of 10