How to Calculate Acceleration and Tension in a Two-Trolley System

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SUMMARY

The discussion focuses on calculating the acceleration and tension in a two-trolley system where Trolley 1 has a mass of 1 kg and Trolley 2 has a mass of 3 kg, with a horizontal force of 12 N applied. The calculated acceleration of the system is 6.85 m/s². The tension in the string connecting the trolleys is determined to be 16.65 N for Trolley 1 and 49.95 N for Trolley 2. The calculations utilize Newton's second law and require separating forces acting in the x and y directions.

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Homework Statement



A student pulls two trolleys, which are connected by a piece of string. the mass of the string is negligible. there is a horizontal force of 12N

Trolley1: 1kg
Trolley 2: 3kg

1. Calculate acceleration of the trolley.
2. determine the force exerted by the connecting string on each trolley.

Homework Equations





The Attempt at a Solution



1. F(res)=ma F(g)=mg
F(g) - F(A)= ma =(4)(9,8)
39.4 - 12= 4a =39,2
a= 6,85m/s^2

2. Trolley1: F(res)=ma F(g)= mg
F(T)-F(g)=ma = (1)(9,8)
F(T) - 9,8= 6,85 =9,8
F(T)= 16,65N

Trolley2: F(res)= ma F(g)= mg
F(T)- F(g)= ma =(3)(9,8)
F(T) - 29,4) = (3)(6,85) =29,4
F(T)= 49,95N
 
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You need to separate out the x and y directions. The applied force of 12 N is horizontal. Look at the force in the x direction (horizontal direction) acting on the system, and use F_net in x direction = ma to solve for the acceleration in the x direction. Don't add F(g) forces to it, since those act in the y direction.

Calculate the tension in the string connecting the trolleys by drawing a free body diagram of the last trolley, identifying the forces, and using Newton 2 again on that body.

Check your work by drawing a FBD of the first trolley and proceeding in the same manner.
 

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