A solid uniform 45.0-kg ball of diameter 32.0 cm is supported against a vertical, frictionless wall by a thin 30.0-cm wire of negligible mass (Fig. P5.65).

(a) Draw a free-body diagram for the ball, and use the diagram to find the tension in the wire.

(b) How hard does the ball push against the wall?

The given data can be listed below as:

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

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 $T\cos \theta$and $T\sin \theta$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 $\theta$can be identified, which will be beneficial for evaluating the tension.

The equation of the angle can be expressed as:

$$\begin{gathered} \mathrm{sin\theta }=\frac{0.16\mathrm{m}}{0.16\mathrm{m}+0.3\mathrm{m}} \\ =\frac{0.16\mathrm{m}}{0.46\mathrm{m}} \\ =20.35° \end{gathered}$$

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:

$$\begin{gathered} \sum _{}{F}_x=0 \\ T\sin \theta -n=0 \\ n=T\sin \theta \end{gathered}$$ (i)

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

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

$\sum _{}{F}_y=0T\cos \theta -mg=0T=\frac{mg}{\cos \theta }$ (ii)

Here $\sum _{}{F}_y$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 $\theta$in the equation (ii).

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

Thus, the tension in the wire is 470 N .

The equation (i) has been recalled below.

$$\begin{gathered} \sum _{}{\mathrm{F}}_{\mathrm{x}}=0 \\ \mathrm{Tsin\theta }-\mathrm{n}=0 \\ \mathrm{n}=\mathrm{Tsin\theta } \end{gathered}$$

Substitute 470 N for T and $20.35°$for$\theta$ in the equation (i).

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

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.