Recent content by kartis

Is it possible for there to be two different forces of tension on opposite ends if the rope is being pulled by forces with different magnitudes?

Thank you for you help!

I think I understand now, thanks. But how will we solve that with the equations? Can I still use Fnet = m*a? Or do I just figure it out by recognizing the forces are unbalanced, so the rope would move in the direction of the unbalanced force?

If they're pushing on the ground, that means the ground may be creating friction helping them each to move to the left or right, but I believe the student with the greater acceleration wins? I feel like there's multiple ways to do this problem that I can't wrap my head around. Lol i guess...

The mass of the rope isn't given. So i assume you can only find the tension after you've calculated the acceleration of the students.

I think I've used the equations from the second law throughout the question. The only thing applied from Newton's third law could be that since every action force has an equal reaction force, the tension would be equal on both ends of the rope.

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...