A diagram shows two masses on a bench top connected by a light inextensible string. The 3 kg mass is on the bench and connected by a string to a 5 kg mass which hangs off the bench... A diagram shows two masses on a bench top connected by a light inextensible string. The 3 kg mass is on the bench and connected by a string to a 5 kg mass which hangs off the bench and accelerates downwards. With an acceleration of 3 m/s^2, find (1) the net force acting on the system, (2) the tension, T, in the string, and (3) the frictional force between the 3 kg mass and the bench top.

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Understand the Problem

The question describes a system of two masses connected by a string over a pulley. One mass (3 kg) rests on a table and experiences friction, while the other mass (5 kg) hangs vertically. The system is accelerating, and we will need to calculate the net force acting on the system, tension in the string, and the frictional force acting on the 3 kg mass.

Answer

(i) $24 N$ (ii) $34 N$ (iii) $25 N$
Answer for screen readers

(i) The net force acting on the system is $24 N$. (ii) The tension in the string is $34 N$. (iii) The frictional force between the 3 kg mass and the bench top is $25 N$.

Steps to Solve

  1. Calculate the net force acting on the system. The net force can be calculated using Newton's second law, $F = ma$, where $m$ is the total mass of the system and $a$ is the acceleration. The system consists of both masses, so we add them together, $m_{total} = 3kg + 5kg = 8kg$. Then we can plug in the values into the equation.

$F_{net} = (8 kg)(3 m/s^2) = 24 N$

  1. Calculate the tension in the string. Consider the 5 kg mass. The forces acting on it are the tension $T$ upwards and the weight $W = mg$ downwards. The net force on this mass is $F = W - T = ma$. We can solve for $T$.

$W = mg = (5 kg)(9.8 m/s^2) = 49 N$ $49 N - T = (5 kg)(3 m/s^2)$ $49 N - T = 15 N$ $T = 49 N - 15 N = 34 N$

  1. Calculate the frictional force. Consider the 3 kg mass. The forces acting on it are the tension $T$ to the right and the frictional force $F_R$ to the left. The net force acting on this mass is $F = T - F_R = ma$.

$34 N - F_R = (3 kg)(3 m/s^2)$ $34 N - F_R = 9 N$ $F_R = 34 N - 9 N = 25 N$

(i) The net force acting on the system is $24 N$. (ii) The tension in the string is $34 N$. (iii) The frictional force between the 3 kg mass and the bench top is $25 N$.

More Information

The net force on the system is different than the net force on each of the individual masses. When calculating the tension, we can isolate one of the masses and analyze the forces acting only on that mass.

Tips

A common mistake is to forget to include both masses when calculating the net force on the system. Also, students often have trouble setting up the correct sign convention for the forces, especially when calculating the tension.

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