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

[joe_]

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?

 

[PhanthomJay]

You're looking at the wrong triangle. The force parallel to the incline is the component of the applied horizontal force.

 

[Moetasim]

I would first commend the student for attempting to solve the problem and using the appropriate equations and trigonometric functions. I would also point out that their approach is correct, as the sum of all forces must be equal to zero since the piano is moving at a constant velocity down the ramp.

In order to calculate the force parallel to the floor, the student should use the same approach as they did for the force parallel to the incline. However, instead of using the sine function, they should use the cosine function, as the force is acting perpendicular to the incline.

Therefore, the correct solution would be:

Cos (11) (1764N) = 1715N parallel to the floor

It is possible that the discrepancy between the student's answer and the answer in the back of the book is due to rounding errors or a different method of calculation. However, the student's approach is correct and they should continue to use it in future problem solving.