small wheeled cart

Question

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 ms^-2 the rope is inelastic, and the frictional force acting upon the cart is 30 N.

A) What is the tension in the rope?

B) What force are you applying to the ground in order to produce this acceleration?

Answer 1

Draw a free-body diagram of the sister:

∑F_y: F_Normal - Weight

∑F_x: Tension - F_Friction

Since there is no acceleration in the y direction, ∑F_y = 0

In the x direction, ∑F_x = ma_x so T - F_Friction = m_sister/cart*a_x

T - 30 = 35*0.10 or T = 3.5 + 30 = 33.5N. However, the given frictional force has only 1 sig fig so doing this properly would require the answer to be reduced to 1 sig fig, which seems a little flaky here, but that makes the tension 30N

Regarding question b. lets draw a free-body of you:

Considering the y direction, ∑F_y: F_Normal - Weight and again since there is no acceleration in the y direction, ∑F_y = 0 therefore F_Normal - Weight = 0 so F_Normal = Weight = m_you * g = 65*9.8 = 637 N however, in reading the question, it says "What force are you applying to the ground in order to produce this acceleration?" This F_Normal is not being used to produce this acceleration because it is not in the direction of the acceleration so I would say there is no force being applied to the ground that produces the acceleration. The normal force keeps you from falling through the ground but it doesn't produce the acceleration. The acceleration is in the x direction and the force on the ground is in the y direction and these are not related to each other.