Question

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?

Answer 1

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

Answer 2

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

Answer 3

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

Answer 4

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

Answer 5

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

Answer 6

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

Answer 7

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 ?