Q65P - University Physics with Modern Physics

Question

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?

Step-by-Step Solution

Answer

(a) 

.

The tension in the wire is 470 N.

(b) The ball push 163 N against the wall.

Step 1: Identification of given data

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 $L=30.0 \mathrm{cm}\times \left(\frac{1\mathrm{m}}{100\mathrm{cm}}\right)=0.3 \mathrm{m}$.

Step 2: Significance of the Newton’s second law

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.

Step 3: (a) Determination of the free-body diagram

The free-body diagram has been drawn below:

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

Step 4: (a) Determination of the tension in the wire

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 x and  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:

 

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

 

Here $\sum _{ }{F}_y$ is the summation of the forces acting in the y 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 .

Step 5: (b) Determination of the force the ball push against the wall

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.