Question

A grand piano with a mass of 345 kg that's supposed to be delivered is pushed up to a delivery truck using an inclined plane elevated at 40°. If the applied force onto the piano is 3530 N, will it be enough to push the grand piano up the ramp? Show your solution for the values of net force along the slope of the incline utilizing both the static and kinetic friction. The coefficient of static friction is 0.52 and the coefficient of kinetic friction 0.15

Answer 1

Let us consider a giant piano of mass $m$. The piano is pushed up by a force $F$ to a delivery truck using an inclined plane elevated at an angle $\theta$ as shown in the figure.

The weight $mg$ of the piano has two components $mg\sin\theta$ and $mg\cos\theta$ as shown in the figure.

The normal force on the inclined plane is $N=mg\cos\theta$

If $\mu$ be the coefficient of friction, the frictional force is $f=\mu N=\mu mg\cos\theta$

Since we are trying to push the piano upward, the force of friction will be downward along the slope of the plane.

As shown in the figure, total downward force along the slope of the plane will be $f+mg\sin\theta$

Since the piano is pushed up by force $F$ , the net upward force will be

$F_{net}=F-(f+mg\sin\theta)\\ \Rightarrow F_{net}=F-(\mu mg\cos\theta+mg\sin\theta)\\ \Rightarrow F_{net}=F-mg(\mu\cos\theta+\sin\theta)$

Initially the body is at rest and the static friction will come into play.

Substituting $F=3530N,m=345 kg, g=9.8m/s^2, \theta = 40\degree, \mu=0.52$ for static friction,

$F_{net}=3530-345\times 9.8(0.52\cos40\degree+\sin40\degree)\\ \Rightarrow F_{net}=9.9N$

Thus, the piano will start to move upward.

Substituting $\mu=0.15$ for kinetic friction,

$F_{net}=3530-345\times 9.8(0.15\cos40\degree+\sin40\degree)\\ \Rightarrow F_{net}=968.2N$

Answer: 3530N will be enough to move the piano up the ramp.

The net force during static friction = 9.9N

The net force during kinetic friction = 968.2N