## anonymous one year ago Trent and Dyson pull on opposite sides of a shopping cart that has a mass of 13kg.Trent pulls the cart to the right with a force of 85N and Dyson pulls the cart to the left with a force of 96 N. A.draw a free body diagram of the shopping cart. B. Write the expression for the net force on the shopping cart along the y-axis C.Write the expression for the net force on the shopping cart along the x-six. D. What is the normal force acting on the shopping cart? E.what is the net force on the shopping cart along the x-axis including direction?

1. johnweldon1993

|dw:1433899313393:dw| Remember $$\large \Sigma F = ma$$ From that FBD we have 2 forces along the y-axis we have 2 forces along the x-axis We can find out the Normal force by using the fact tht the force of gravity is mass x acceleration due to gravity And the net force in the horizontal direction can be solved by just combining the vector

2. anonymous

For the normal force I got 10 to the left 95-85

3. anonymous

For b. Finding the net force would I multiply 13*85

4. johnweldon1993

Normal force would oppose the force of gravity $$\large \Sigma_y = 0$$ since the cart is not accelerating in the y-direction $\large 0 = -mg + N = -(13kg\times 9.81 \frac{m}{s^2}) + N$ $\large N = 13kg \times 9.81\frac{m}{s^2} = ?$ Same process with the x-direction

5. anonymous

so would the normal force be 127.4 and would the net force be 10 to the left.

6. johnweldon1993

Your Normal Force (I'm assuming you used 9.8 instead of 9.81) is correct...in Newtons of course As far as the net force in the x-direction, not quite $\large [-96N\hat i + 85N \hat i] = -11N\hat i$ so it would be 11N to the neft

7. anonymous

ohhh okay , and for the acceleration of the shopping cart i added 96 and 85 and got 181 and then i divided it by 13 and got 13.92 is that correct ?