Sure, let's solve this step-by-step.
### Given:
- Mass of the grocery cart ([tex]\( m \)[/tex]) = 7.32 kg
- Force applied ([tex]\( F \)[/tex]) = 14.7 N
- Angle below horizontal ([tex]\( \theta \)[/tex]) = -32.7°
### Steps:
1. Calculate the gravitational force (weight) acting on the cart:
[tex]\[
F_g = m \times g
\][/tex]
where [tex]\( g \)[/tex] (acceleration due to gravity) is 9.81 m/s².
So,
[tex]\[
F_g = 7.32 \, \text{kg} \times 9.81 \, \text{m/s}^2 = 71.8092 \, \text{N}
\][/tex]
2. Determine the vertical component of the pushing force:
The vertical component of the force ([tex]\( F_{\text{vertical}} \)[/tex]) can be found using the sine of the angle because the force is below the horizontal:
[tex]\[
F_{\text{vertical}} = F \times \sin(\theta)
\][/tex]
Given that [tex]\( \theta = -32.7^\circ \)[/tex],
[tex]\[
\sin(-32.7^\circ) = -\sin(32.7^\circ)
\][/tex]
So,
[tex]\[
F_{\text{vertical}} = 14.7 \, \text{N} \times \sin(-32.7^\circ)
\][/tex]
[tex]\[
F_{\text{vertical}} = 14.7 \, \text{N} \times -0.540640817 = -7.941532711
\][/tex]
Hence, the vertical component of the pushing force ([tex]\( F_{\text{vertical}} \)[/tex]) is approximately -7.94 N, which acts downward.
3. Determine the normal force ([tex]\( n \)[/tex]):
The normal force is the force exerted by the ground to support the cart vertically. It is the reactive force to the sum of the gravitational force and the vertical component of the applied force.
[tex]\[
n = F_g + F_{\text{vertical}}
\][/tex]
Here,
[tex]\[
n = 71.8092 \, \text{N} + (-7.941532711 \, \text{N}) = 63.867667288978474 \, \text{N}
\][/tex]
So, the total normal force acting on the cart is approximately 63.87 N.
### Answer:
[tex]\[
n = 63.87 \, \text{N}
\][/tex]
Thus, the total normal force acting on the grocery cart is [tex]\( 63.87 \, \text{N} \)[/tex].
