91Ó°ÊÓ

The given data can be listed below as:

• The mass of the uniform ball is m .
• The angle of the ball with the horizontal is$\mathrm{θ}=35.0°$.

Tension is mainly used for maintaining the integrity of a particular system. Moreover, tension is also used for pulling equal energy of a particular system from both the ends of the system.

The free-body diagram of the ball has been described below:

Here, in this free body diagram, the ball exerts a force mg on the ground. The two components of the force mg are mg sin$\mathrm{θ}$ and$mg\cos \mathrm{θ}$ respectively. The ball is also exerting a tension T that also has two components such as T sin$\mathrm{θ}$ and$T\cos \mathrm{θ}$ respectively. The ground is also exerting a reaction force R on the ball.

According to the diagram, the summation of the forces in the x and also in the y direction is zero.

The equation of the summation of the forces in the x direction is expressed as:

$\underset{}{∑}{F}_x=0$

Here$\underset{}{∑}{F}_x$is the summation of the forces in the x direction.

The above equation can also be expressed as:

$R=mg\cos \mathrm{θ}+T\sin \mathrm{θ}$ (i)

The equation of the summation of the forces in the direction is expressed as:

$\underset{}{∑}{F}_y$= 0

Here$\underset{}{∑}{F}_y$is the summation of the forces in the y direction.

The above equation can also be expressed as:

$$\begin{gathered} \mathrm{°Õ³¦´Ç²õθ}=\mathrm{³¾²\mu ²õ¾\pm ²Ôθ} \\ \mathrm{T}=\mathrm{³¾²\mu ³Ù²¹²Ôθ} \end{gathered}$$ (ii)

Substitute$mg\tan \mathrm{θ}$for T in the equation (i), and we get,

$R=mg\cos \mathrm{θ}+mg\tan \mathrm{θ}\sin \mathrm{θ}$

Substitute$35.0°$for$\mathrm{θ}$in the above equation, and we get,

$$\begin{gathered} \mathrm{R}=\mathrm{mgcos}35.0°+\mathrm{mgtan}35.0°\sin 35.0° \\ =\mathrm{mg}\left(0.819+0.7×0.57\right) \\ =1.22\mathrm{mg} \end{gathered}$$

Thus, the surface of the ramp has to push force of 1.22 mg on the wall.

The equation (ii) is recalled again.

$$\begin{gathered} \mathrm{°Õ³¦´Ç²õθ}=\mathrm{³¾²\mu ²õ¾\pm ²Ôθ} \\ \mathrm{T}=\mathrm{³¾²\mu ³Ù²¹²Ôθ} \end{gathered}$$

Substitute$35.0°$ for$\mathrm{θ}$ in the above equation, and we get,

$$\begin{gathered} \mathrm{T}=\mathrm{mgtan}35.0° \\ =0.7\mathrm{mg} \end{gathered}$$

Thus, the tension in the wire is 0.7mg .