# Supposedly Simple Tension Problem

**Q: You are pulling your younger sister along in a small wheeled cart. You weigh 65.0 kg and the combined mass of your sister and the cart is 35.0 kg. You are pulling the cart via a short rope which you pull horizontally. You hold one end of the rope and your sister holds the other end. If you are accelerating at a rate of 0.10 m/s-2, the rope is inelastic, and the frictional force acting on the cart is 30 N: what is the tension on the rope?**

Now, I’ve tried to search for this online, and all answers I’ve found go something like this:

F = MA = Tension – friction

Thus MA = Tension – friction

T = MA + F = 35*.1 + 30 = 33.5 N

Is this correct? Assuming from the equation above (Tension – friction), tension would be in the opposite direction of the friction (which I assume would be opposite the direction of motion). I’m a bit confused by this, as I thought tension was supposed to be opposite of the applied force.

Please explain how to do this problem clearly ðŸ™‚ Thanks!

http://ift.tt/MwJSDh

## Leave a comment