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A shopper pushes a 7.32 kg grocery cart
with a 14.7 N force directed at
-32.7° below the horizontal.
What is the normal force on the cart?

Respuesta :

skyluke89

The normal force on the cart is 79.7 N

Explanation:

In order to find the normal force, we have to analyze the forces acting on the cart on the vertical direction.

In the vertical direction, we have the following forces:

  • The weight of the cart, downward, of magnitude [tex]mg[/tex], where m is the mass of the cart and g is the acceleration of gravity
  • The normal force on the cart, upward, we indicate it with N
  • The component of the pushing force acting in the vertical direction, downward, of magnitude [tex]F sin \theta[/tex], where F is the magnitude of the force and [tex]\theta[/tex] is the angle of the force with the horizontal

Therefore, the equation of the forces on the cart in the vertical direction is:

[tex]N - mg - F sin \theta = 0[/tex]

where the net force is zero since the cart is balanced in the vertical direction. We have:

[tex]m=7.32 kg\\g=9.8 m/s^2\\F=14.7 N\\\theta =32.7^{\circ}[/tex]

We take the angle as positive since we are already considering the downward direction in the equation.

Substituting and solving for N, we find the normal force:

[tex]N=mg+F sin \theta = (7.32)(9.8) + (14.7)(sin 32.7^{\circ})=79.7 N[/tex]

Learn more about forces:

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