How Does Newton's Laws Apply to a Piano on an Incline?

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SUMMARY

This discussion centers on applying Newton's Laws to calculate the force required to push a piano down an incline. The piano, with a mass of 180 kg, slides at a constant velocity down a ramp inclined at 11 degrees. The force exerted by the man pushing parallel to the incline is calculated to be 336.6 N, while the force applied parallel to the floor is determined to be 343 N, correcting the initial misunderstanding regarding the triangle used for calculations.

PREREQUISITES
  • Understanding of Newton's Laws of Motion
  • Basic knowledge of trigonometric functions (SOHCAHTOA)
  • Familiarity with force calculations (F = ma)
  • Concept of components of forces in inclined planes
NEXT STEPS
  • Study the application of Newton's Laws in various physical scenarios
  • Learn about resolving forces into components on inclined planes
  • Explore the effects of friction on motion in inclined systems
  • Investigate the relationship between mass, weight, and gravitational force
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Students studying physics, particularly those focusing on mechanics and force analysis, as well as educators seeking to clarify concepts related to inclined planes and Newton's Laws.

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Homework Statement


A man pushes on a piano with mass 180 kg so that it slides at constant velocity down a ramp that is inclined at 11deg above the horizontal floor. Neglect any friction acting on the piano. Calculate the magnitude of the force applied by the man if he pushes (a) parallel to the incline and (b) parallel to the floor


Homework Equations


Trig (SOHCAHTOA)
F= ma


The Attempt at a Solution


At first I thought this would be zero because the acceleration is 0. But looking at the back of the book the answer is NOT zero, so I attempted to figure out a way to solve it..

I got the first part right, but can't figure out the second part (b)

f=ma
f=(180kg)(9.8m/s^2)
f=1764N

Sin (11) (1764N) = 336.6 N parallel to the ramp (answer checked out in the back of the book)

Now, for the second part... I know that there is zero acceleration so the sum of all forces must be zero. So I know from the first part the force is 336.6, so the opposite must also be 336.6. I also know that it makes a 11 deg angle with the horizontal component (what I am looking for) due to it being a similar triangle. So:

Cos (11) (336.6N)= 330.4N. (which is apparently wrong, my book says it should work out to 343N. How so?
 
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You're looking at the wrong triangle. The force parallel to the incline is the component of the applied horizontal force.
 

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