Calculating Work and Forces on a Sliding Piano

[VanKwisH]

Power,Work,Energy problem...

Homework Statement

A 300 kg piano slides 4.5 m down a 25degree incline and is kept from accelerating
by a man who is pushing back on it parallel to the incline the effective co-efficient of friction
is 0.4 Calculate
A) the force exerted by the man (answer is 180N)
B) the work done by the man on the piano (Answer is -800 J)
C) the work done by the Friction force, (Answer is -4800 J)
D) the work done by the force of gravity(Answer is 5600J)
E) the net work done on the piano(answer is 0)

Homework Equations

Fgx = sin25 x Fg
Fgy = cos25 x Fg
Ff=uFn
W=FxD
Fnet=Fa + Ff + Fgx(or Fgy)

The Attempt at a Solution

D = 4.5
mu = 0.4
Fg=2940
Fgx= -389.1
Fgy= -2914 therefore Fn= +2914
Ff = mu*Fn = -1166 N

I got the some info but... but i ain't gettin the answers ... anyone know??

 

[Astronuc]

One needs to show some work here.

Start with

Fgx = sin25 x Fg
Fgy = cos25 x Fg
Ff=uFn

and enter the appropriate numbers.

There is a weight (mg) component pointing down the ramp. At the same time, friction is acting up the ramp, so the friction is helping the man by reducing the force that he is required to resist the sliding piano.

What is the force of the piano pointing down the ramp? What is the friction force pointing up the ramp? What is the difference - the force the man must apply?

 

[VanKwisH]

One needs to show some work here.

Start with

Fgx = sin25 x Fg
Fgy = cos25 x Fg
Ff=uFn

and enter the appropriate numbers.

There is a weight (mg) component pointing down the ramp. At the same time, friction is acting up the ramp, so the friction is helping the man by reducing the force that he is required to resist the sliding piano.

What is the force of the piano pointing down the ramp? What is the friction force pointing up the ramp? What is the difference - the force the man must apply?

D = 4.5
mu = 0.4
Fg=2940
Fgx= -389.1
Fgy= -2914 therefore Fn= +2914
Ff = mu*Fn = +1166 N


so then Fa = Ff + Fgx??
which means Fa = 1166 - 389.1? = 776.9??
but that doesn't get me 180 N

 

[VanKwisH]

O wow nvm my calculator was in radian mode srry nvm

 

[Astronuc]

mg = 300 kg * 9.81 m/s² = 2943 N

Now of that, mg sin (25°) is pointing down the ramp.

2943 N sin (25°) = 2943 N * 0.423 = 1244 N

Pointing up the ramp is the normal component of the weight (mg cos (25°)) and the friction force is \mu = 0.4 times the normal force.

Then take the difference in magnitude (or add vectorally) and determine the force that the man must apply to achieve a zero net force to maintain constant velocity (or zero acceleration).

 

[Astronuc]

O wow nvm my calculator was in radian mode srry nvm

It happens. One has to be careful with that, and the units.

It's best to write equations with units in order to avoid missing a conversion, especially when using mixed units, e.g. British with MKS or cgs.

 

[VanKwisH]

haha thanks for explaining it anyways :D