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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?

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  1. johnweldon1993
    • one year ago
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    |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
    • one year ago
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    For the normal force I got 10 to the left 95-85

  3. anonymous
    • one year ago
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    For b. Finding the net force would I multiply 13*85

  4. johnweldon1993
    • one year ago
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    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
    • one year ago
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    so would the normal force be 127.4 and would the net force be 10 to the left.

  6. johnweldon1993
    • one year ago
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    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
    • one year ago
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    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 ?

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