Pushing a Box On A Plane 30deg Plane Help?

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SUMMARY

The discussion focuses on analyzing the forces acting on a box being pushed on a 30-degree inclined plane. Key calculations include the weight of the box, which is determined using the formula M*G, yielding 544.455 N. The parallel force is calculated as 272.228 N using sin(30), while the perpendicular force is 471.512 N using cos(30). The man applies a pushing force of 314.341 N, derived from tan(30). The confusion arises regarding the normal force, which must account for both the gravitational component and the pushing force's vertical component.

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  • Understanding of Newton's Second Law (F=MA)
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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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