91Ó°ÊÓ

Given Data:

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

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.

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

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

Here $\mathrm{θ}$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,

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

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

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

$mg\sin \mathrm{θ}=F_1\cos \mathrm{θ}$

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,

$$\begin{gathered} \left(180\mathrm{kg}\right)×\left(9.8\mathrm{m}/{\mathrm{s}}^2\right)×\left(\sin {19}^{°}\right)=F_1\left(\cos {19}^{°}\right) \\ {F}_1=607\mathrm{N} \end{gathered}$$

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