Question

A shopper pushes a 7.32 kg grocery cart

Answer 1

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 $mg$, 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 $F sin \theta$, where F is the magnitude of the force and $\theta$ is the angle of the force with the horizontal

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

$N - mg - F sin \theta = 0$

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

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

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:

$N=mg+F sin \theta = (7.32)(9.8) + (14.7)(sin 32.7^{\circ})=79.7 N$

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