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Sagot :
To determine the net force in the [tex]\( y \)[/tex]-direction for a box being pushed down at an angle of 32 degrees on a rough surface, we need to consider all the forces acting in the vertical direction.
1. Normal Force [tex]\( F_N \)[/tex]:
- The normal force acts upward perpendicular to the surface.
2. Gravitational Force [tex]\( F_g \)[/tex]:
- The gravitational force acts downward and is equal to [tex]\( mg \)[/tex], where [tex]\( m \)[/tex] is the mass of the box and [tex]\( g \)[/tex] is the acceleration due to gravity.
3. Vertical Component of the Pushing Force [tex]\( F_p \)[/tex]:
- The pushing force makes an angle of 32 degrees with the horizontal. Therefore, it has a vertical component as well. This component can be found by using the sine of the angle:
[tex]\[ F_{p, \text{vertical}} = F_p \sin(32^\circ) \][/tex]
- Since the box is pushed downwards, the vertical component of the pushing force acts downward.
To find the net force in the [tex]\( y \)[/tex]-direction, we need to combine these forces. The net force ([tex]\( F_{\text{net}, y} \)[/tex]) will be the sum of these vertical forces. Considering their directions:
- The normal force [tex]\( F_N \)[/tex] acts upward (positive direction).
- The gravitational force [tex]\( F_g \)[/tex] acts downward (negative direction).
- The vertical component of the pushing force, [tex]\( F_p \sin(32^\circ) \)[/tex], also acts downward (negative direction).
Thus, the equation for the net force in the [tex]\( y \)[/tex]-direction is:
[tex]\[ F_{\text{net}, y} = F_N - F_g - F_p \sin(32^\circ) \][/tex]
Therefore, the correct choice is:
[tex]\[ F_{\text{net}, y} = F_{N} - F_g - F_p \sin(32^\circ) \][/tex]
Hence, the fourth option is the correct one:
[tex]\[ \boxed{4} \][/tex]
1. Normal Force [tex]\( F_N \)[/tex]:
- The normal force acts upward perpendicular to the surface.
2. Gravitational Force [tex]\( F_g \)[/tex]:
- The gravitational force acts downward and is equal to [tex]\( mg \)[/tex], where [tex]\( m \)[/tex] is the mass of the box and [tex]\( g \)[/tex] is the acceleration due to gravity.
3. Vertical Component of the Pushing Force [tex]\( F_p \)[/tex]:
- The pushing force makes an angle of 32 degrees with the horizontal. Therefore, it has a vertical component as well. This component can be found by using the sine of the angle:
[tex]\[ F_{p, \text{vertical}} = F_p \sin(32^\circ) \][/tex]
- Since the box is pushed downwards, the vertical component of the pushing force acts downward.
To find the net force in the [tex]\( y \)[/tex]-direction, we need to combine these forces. The net force ([tex]\( F_{\text{net}, y} \)[/tex]) will be the sum of these vertical forces. Considering their directions:
- The normal force [tex]\( F_N \)[/tex] acts upward (positive direction).
- The gravitational force [tex]\( F_g \)[/tex] acts downward (negative direction).
- The vertical component of the pushing force, [tex]\( F_p \sin(32^\circ) \)[/tex], also acts downward (negative direction).
Thus, the equation for the net force in the [tex]\( y \)[/tex]-direction is:
[tex]\[ F_{\text{net}, y} = F_N - F_g - F_p \sin(32^\circ) \][/tex]
Therefore, the correct choice is:
[tex]\[ F_{\text{net}, y} = F_{N} - F_g - F_p \sin(32^\circ) \][/tex]
Hence, the fourth option is the correct one:
[tex]\[ \boxed{4} \][/tex]
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