Complete Question
The complete question is shown on the
Answer:
The ascending order would be 2nd < 1st < 3rd
Explanation:
Generally the the Normal force is mathematically represented as
[tex]Normal \ Force = Fsin \theta[/tex]
=> [tex]Normal \ Force \ \alpha \ sin \theta[/tex]
For the first drawing the value [tex]\theta[/tex] is between that of the the second and the third drawing so the Normal force would also be between the normal forces of the second and the third drawing
For the second drawing whose value of [tex]\theta[/tex] is less than that of the first and the third the normal force would also be less than that of the first and third
For the the third drawing whose value is (90°) which is higher than the first and the second the normal force would also be higher than the first and the second
Answer:
drawing 2< drawing 1 < drawing 3
Explanation:
Complete Question is:
The drawings(attachment) show three examples of the force with which someone pushes against a vertical wall. In each case the magnitude of the pushing force is the same. Rank the normal forces that the wall applies to the pusher in ascending order (smallest first)
1st drawing is a medium angle
2nd drawing is a small angle
3rd drawing is a large angle (almost 90 degrees and perpendicular to the wall)
Normal Force= Fsinθ
Here is θ is the angle between the wall and the force applied that is F.
As θ increases, value of sinθ increases. When θ is 90, sinθ is maximum that is 1 and value of resultant.
Therefore, normal force is greatest in diagram 3 and least in diagram 2
