91Ó°ÊÓ

The given data can be listed below as:

• The mass of the ball is m =45.0 kg .
• The diameter of the ball is$d=32.0\mathrm{cm}×\left(\frac{1\mathrm{m}}{100\mathrm{cm}}\right)=0.32\mathrm{m}$.
• The radius of the ball is $R=\frac{\mathrm{d}}{2}=\frac{\mathrm{d}}{2}=\frac{0.32\mathrm{m}}{2}=0.16\mathrm{m}$.
• The length of the wire is $L=30.0\mathrm{cm}×\left(\frac{1\mathrm{m}}{100\mathrm{cm}}\right)=0.3\mathrm{m}$.

Newton’s second law states that acceleration occurs when a particular force acts on an object. The force is equal to the product of acceleration and mass.

The free-body diagram has been drawn below:

From the above diagrams, it has been observed that the wire exerts a tension Ton the ball, the weight of the ball is mg, and the force exerted by the wall isn. The tension has two components such as $T\cos \mathrm{θ}$and $T\sin \mathrm{θ}$in the horizontal and in the vertical directions, respectively.

The diagram to find the tension in the wire has been drawn below:

From the above diagram, the value of the angle $\mathrm{θ}$can be identified, which will be beneficial for evaluating the tension.

The equation of the angle can be expressed as:

$$\begin{gathered} \mathrm{²õ¾\pm ²Ôθ}=\frac{0.16\mathrm{m}}{0.16\mathrm{m}+0.3\mathrm{m}} \\ =\frac{0.16\mathrm{m}}{0.46\mathrm{m}} \\ =20.35° \end{gathered}$$

From the free-body diagram, the summation of all the forces in xand in the y direction is zero.

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

$$\begin{gathered} \underset{}{∑}{F}_x=0 \\ T\sin \mathrm{θ}-n=0 \\ n=T\sin \mathrm{θ} \end{gathered}$$ (i)

Here $\underset{}{∑}{F}_x$ is the summation of the forces acting in the x direction, n is the normal force, T is the tension and $\mathrm{θ}$is the angle subtended with the wall.

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

$\underset{}{∑}{F}_y=0T\cos \mathrm{θ}-mg=0T=\frac{mg}{\cos \mathrm{θ}}$ (ii)

Here $\underset{}{∑}{F}_y$is the summation of the forces acting in they direction, n is the normal force, is the tension and is the angle subtended with the wall.

Substitute 45.0 kg for m , $9.8\mathrm{m}/{\mathrm{s}}^2$for T, and $20.35°$for $\mathrm{θ}$in the equation (ii).

$$\begin{gathered} \mathrm{T}=\frac{\left(45.0\mathrm{kg}\right)\left(9.8\mathrm{m}/{\mathrm{s}}^2\right)}{\cos 20.35°} \\ =\frac{441\mathrm{kg}.\mathrm{m}/{\mathrm{s}}^2}{0.93} \\ =470\mathrm{kg}.\mathrm{m}/{\mathrm{s}}^2×\left(\frac{1\mathrm{N}}{1\mathrm{kg}.\mathrm{m}/{\mathrm{s}}^2}\right) \\ =470\mathrm{N} \end{gathered}$$

Thus, the tension in the wire is 470 N .

The equation (i) has been recalled below.

$$\begin{gathered} \underset{}{∑}{\mathrm{F}}_{\mathrm{x}}=0 \\ \mathrm{T²õ¾\pm ²Ôθ}-\mathrm{n}=0 \\ \mathrm{n}=\mathrm{T²õ¾\pm ²Ôθ} \end{gathered}$$

Substitute 470 N for T and $20.35°$for$\mathrm{θ}$ in the equation (i).

$$\begin{gathered} \mathrm{n}=\left(470\mathrm{N}\right)\sin 20.35° \\ =\left(470\mathrm{N}\right)×0.347 \\ =163\mathrm{N} \end{gathered}$$

According to the third law of Newton, the force exerted by the wall is equal to the magnitude of the force exerted by the ball.

Thus, the ball push 163 N against the wall.