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Homework Statement:

Two students are having a tug of war. A 60kg student pulls on a rope with a force of 40N to the left. A 70kg student pulls on the other end of the rope with a force of 50N to the right. Find the
a) acceleration of the two students
b) force of tension in the rope
Relevant Equations:
 Newton's Third Law
a) I think you can just use Fnet = m*a, so for
student 1: a = 40N/60kg → a = 0.667 m/s^2
student 2: a = 50N/70kg → a = 0.714 m/s^2
b) Fnet = F  T, rearrange to solve for tension, → T = F  ma
Student 1, T = 40N  (60kg*0.667m/s^2) → T = 0.02N
Student 2, T = 50N  (70kg*0.71m/s^2) → T = 0.02N
The answer seems wrong since its too small. From my understanding the force of tension of the rope needs to be equal on both sides of the rope. But I'm not sure if that also applies to different applied forces on both ends. The answers to the question aren't given.
student 1: a = 40N/60kg → a = 0.667 m/s^2
student 2: a = 50N/70kg → a = 0.714 m/s^2
b) Fnet = F  T, rearrange to solve for tension, → T = F  ma
Student 1, T = 40N  (60kg*0.667m/s^2) → T = 0.02N
Student 2, T = 50N  (70kg*0.71m/s^2) → T = 0.02N
The answer seems wrong since its too small. From my understanding the force of tension of the rope needs to be equal on both sides of the rope. But I'm not sure if that also applies to different applied forces on both ends. The answers to the question aren't given.