Q9E - University Physics with Modern Physics

Question

A man pushes on a piano with mass 180 kg; it slides at constant velocity down a ramp that is inclined at 19.0° 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.

Step-by-Step Solution

Answer

(a) The magnitude of force parallel to incline is 574 N .

(b) The magnitude of force parallel to the floor is 607 N .

Step 1: Magnitude of Force

Given Data:

• The mass of piano, $m=180\mathrm{kg}$ .
• The inclination of the ramp, $\theta ={90}^{°}$ .

 

The magnitude of Force:

The magnitude of force parallel to the incline can be found by considering the equilibrium of the piano along the horizontal direction. The magnitude of force parallel to the floor can be found by equating the horizontal component of the weight of force to the horizontal component of push force.

Step 2: Determine the magnitude of force parallel to incline (a)

The equilibrium equation for piano for parallel force on the incline is given as:

 $$\begin{gathered} F-mg \sin \theta =0 \\ F=mg \sin \theta \end{gathered}$$ 

 

Here $\theta$ is the angle of the inclined rope, $F$ which is the force applied parallel to the incline.

 

Substitute all the values in the above equation, and we get,

 $$\begin{gathered} F=\left(180\mathrm{kg}\right)\times \left(9.8\mathrm{m}/{\mathrm{s}}^2\right)\times \left(\sin {19}^{°}\right) \\ F=574 \mathrm{N} \end{gathered}$$ 

 

Therefore, the magnitude of force parallel to incline is 574 N.

Step 3: Determine the magnitude of force parallel to the floor (b)

The equilibrium equation for piano for parallel force on the incline is given as:

$mg \sin \theta =F_1\cos \theta$

Here $m$ is the mass of the piano and $F_1$ is the applied force parallel to the floor.

Substitute all the values in the above equation (1), and we get,

$$\begin{gathered} \left(180\mathrm{kg}\right)\times \left(9.8\mathrm{m}/{\mathrm{s}}^2\right)\times \left(\sin {19}^{°}\right)=F_1\left(\cos {19}^{°}\right) \\ F_1=607 \mathrm{N} \end{gathered}$$  

Therefore, the magnitude of force parallel to the floor is 607 N .