Question

Three blocks m1 = 1 kg, m2 = 2 kg, and m3 = 3 kg are connected through a frictionless pulley as shown in the figure below. A downward force F = 48 N is acting on block m1. Using the system approach, determine the magnitude of the acceleration of these blocks.

Answer 1

Consider the system formed by the blocks m1, m2 and m3.

Let the forces acting in the direction of the force F be positive.

There is a total of 4 forces acting on the system: the force F and the weighs of blocks m1, m2 and m3.

The force F and the weight of m1 are positive, while the weighs of blocks m2 and m3 are negative.

The weighs of blocks m1, m2 and m3 can be found by multiplying the masses m1, m2 and m3 by the acceleration of gravity, g=9.81m/s^2:

`\begin{gathered} W_1=m_1g=(1kg)(9.81(m)/(s^2))=9.81N \\ W_2=-m_2g=-(2kg)(9.81(m)/(s^2))=-19.62N \\ W_3=-m_3g=-(3kg)(9.81(m)/(s^2))=-29.43N \end{gathered}`

The net force acting on the system is:

`\begin{gathered} \Sigma F=F+W_1+W_2+W_3 \\ =48N+9.81N-19.62N-29.43N \\ =8.76N \end{gathered}`

The total mass of the system is:

`M=m_1+m_2+m_3=1kg+2kg+3kg=6kg`

According to Newton's Second Law of Motion, the acceleration of the system is given by:

`a=(\Sigma F)/(M)=(8.76N)/(6kg)=1.46(m)/(s^2)`

Notice that none of the options coincides with this answer. Nevertheless, the option 1.33m/s^2 is closest to 1.46m/s^2. We can justify that answer if we consider g=10m/s^2, which leads us to the following calculations:

`\begin{gathered} W_1=(1kg)(10(m)/(s^2))=10N \\ W_2=-(2kg)(10(m)/(s^2))=-20N \\ W_1=-(3kg)(10(m)/(s^2))=-30N \\ \\ \Sigma F=48N+10N-20N-30N=8N \\ \\ a=(8N)/(6kg)=1.33(m)/(s^2) \end{gathered}`

Therefore, the correct choice is: 1.33 m/s^2.