Pushing a Box On A Plane 30deg Plane Help?

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The discussion focuses on calculating forces acting on a box being pushed on a 30-degree incline. The weight of the box is calculated using the formula F=MA, yielding a weight of 544.455 N. The user correctly identifies the parallel and perpendicular forces, but struggles with the normal force, mistakenly believing it should equal the perpendicular force. It is clarified that the man’s pushing force also has two components, which must be considered in the normal force calculation. The conversation emphasizes the need to balance both the gravitational and applied forces acting on the box.
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Homework Statement



Image2.jpg


Homework Equations



F=MA

The Attempt at a Solution



Weight
M*G=55.5*9.81

Parallel force
sin(30)544.455=272.228

Perpendicular force To Plane
cos(30)544.455=471.512

Figured out that the man is pushing with a force of
tan(30)544.455=314.341N

Normal Force
Wouldn't it be the opposite to the perpendicular force? 471.512? But this is wrong? HELP. What other force in the perpendicular direction am I missing?
 
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raven2783 said:

Homework Statement



Homework Equations



F=MA

The Attempt at a Solution



Weight
M*G=55.5*9.81

Parallel force
sin(30)544.455=272.228

Perpendicular force To Plane
cos(30)544.455=471.512

Figured out that the man is pushing with a force of
tan(30)544.455=314.341N

Normal Force
Wouldn't it be the opposite to the perpendicular force? 471.512? But this is wrong? HELP. What other force in the perpendicular direction am I missing?

OK you have the weight of the box perpendicular and parallel ok.

Now the man is pushing with a force that has 2 components too. Since it is at rest, you correctly balanced the downward plane force of the box with the upward plane force from the man and part b) looks correct.

For the last one there are two normal forces to the ramp. The gravity related component added to the Sin30 component of the man pushing .
 

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