How to Calculate Tension in a Cord Around a Pulley

The tension in a cord equals the force it applies at both ends, which, by Newton’s third law, must be equal. If the cord is static, the calculation of the tension is relatively simple. If the forces pulling at the ends are not equal, then the calculation of tension becomes more complicated.

Instructions

1. Calculate the tension in a line draped over a pulley such that the line’s ends suspend a weight equal to 10 Newtons of force. (A Newton (N) is the SI unit for force.) Each end of the line must support the weight equally; otherwise, the line would start moving toward the side being pulled more and stop only after achieving equilibrium. So each end is pulling up 5N. Through the line, one end pulls the other with 5N of force and the other pulls back 5N. The line therefore has a tension of 5N.

2. Calculate the tension in a line if the weights the two ends of the line suspend are not equal—say, of 5N and 3N. Draw the forces affecting the two bodies as vectors. The 3N body has a downward force of 3N and an upward force of tension T. Likewise, the 5N body has a downward force of 5N and an upward force of tension T. Note that the upward force on the 3N body is not 5N, because that’s what the tension would be if both bodies weighed 5N. Since one weighs 3N, there’s less force on the cord, so the tension must be less than 5N. A similar argument shows that the tension has to be more than 3N.

3. Set up the formula F=ma for the 3N body. ma=T-3N. Since m=3N/g, where g is the gravitational acceleration constant 9.8m/s^2, you have m=0.306 kilograms (kg) as the mass of the 3N body. Likewise, the equation and mass for the 5N body is ma=5N-T, with m=0.510kg. So the two equations are 0.306kgxa=T-3N and 0.510kgxa=5N-T.

4. Notice that the acceleration a is the same for both bodies, as well as the cord. They accelerate toward the 5N body’s side together. Since T was subtracted in one equation and added in the other, even the sign of a is equal between equations. Therefore, you can eliminate a from the above equations and combine them to get (T-3N)/0.306kg = (5N-T)/0.510kg. Solving for T gives you T=3.75N, which is between 3N and 5N, as predicted in Step 2.

Tips & Warnings

• There is a simple formula for the tension T in the above setup (which is called an Atwood machine or Atwood's machine). If m1 and m2 are the masses of the two bodies, then T = 2gxm1xm2/(m1+m2). (As above, g is gravitational acceleration.)

References

• Johns Hopkins University: Tension in a String - Atwood’s Machine
• Georgia State University: Atwood’s Machine