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Trent and Dyson pull on opposite sides of a shopping cart that has a mass of 13 kg. Trent pulls the cart to the right with a force of 85 N, and Dyson pulls the cart to the left with a force of 96 N. (Assume there is no friction)

a. Draw the free-body diagram of the shopping cart. (You do not need to draw it perfectly to scale. Just make sure the directions are correct.)

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-axis.

d. What si the normal force acting on the shopping cart?

e. What is the net force on the shopping cart along the x-axis, including direction?

f. What is the acceleration of the shopping cart, including direction?

Respuesta :

b) N=mg

c) F.x=F.T-F.D

d) N=13*9.8=127.4 N

e) F=85-96=-11 N

f) Newton's second law says a=F/m=-11/13=-0.85 N. Negative acceleration means that the cart will go to the left.

Ver imagen ugmichail

Answer:

Part b)

[tex]F_{net} = N - mg = 0[/tex]

Part c)

[tex]F_t - T_d = ma[/tex]

Part d)

N = 127.4 N

Part e)

[tex]F_x = - 11\hat i[/tex]

Part f)

[tex]a = -0.85 \hat i[/tex]

Explanation:

Part a)

FBD for the given block is as following

here normal force is counter balanced by weight while in x direction it will have some net external force

Part b)

For Y direction object is not accelerating

So here the force is balanced

[tex]F_{net} = N - mg = 0[/tex]

Part c)

For X direction all the forces are due to Trent and Dyson

So here the forces are given as

[tex]F_t - F_d = F_{net}[/tex]

Part d)

Form the equation of net force along Y direction we have

[tex]N - mg = 0[/tex]

[tex]N = mg[/tex]

[tex]N = (13)(9.81) = 127.5 N[/tex]

Part e)

From the above equation of net force along x direction we will have

[tex]F_x = F_t - F_d[/tex]

[tex]F_x = 85 - 96[/tex]

[tex]F_x = -11 \hat i N[/tex]

Part f)

From net force in x direction we know

[tex]F_x = ma_x[/tex]

[tex]-11\hat i = 13 a_x[/tex]

[tex]a_x = - 0.85 \hat i[/tex]

Ver imagen aristocles

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