Question

Two racquetballs, each having a mass of 170 g, are placed in a glass jar as shown in Figure P12.59. Their centers lie on a straight line that makes a ${\mathbf{45}}^{\mathbf{°}}$angle with the horizontal.

(a) Assume the walls are frictionless and determine ${\mathit{P}}_{\mathbf{1}}\mathbf{,}{\mathit{P}}_{\mathbf{2}}\mathbf{,}\mathit{a}\mathit{n}\mathit{d}{\mathit{P}}_{\mathbf{3}}$.

(b) Determine the magnitude of the force exerted by the left ball on the right ball..

Short answer

$\begin{array}{l}a)1.67N,3.33N,1.67N\\ b)2.36N\end{array}$

Step-by-step solution

Reaction forces on the wall

(a)

Given,

$\begin{array}{c}{m}_1=m_2\\ =m\\ =170g\\ =170\times {10}^{-3}kg\end{array}$

Now, we need to draw the force diagram of the two balls.

We choose the system to be the two balls (as one unit ) so we did draw, only, the external forces exerted on this system, as you see below.

We know that the system is in static equilibrium, this means that

$\begin{array}{c}\sum F_x=0\\ \sum F_y=0\\ \sum \tau =0\end{array}$...........(1)

Hence, from the figure shown below:

$\begin{array}{l}{P}_3-P_1=0\\ P_3=P_1\end{array}$

And,

$\begin{array}{c}{P}_2-mg-mg=0\\ =2mg\\ =2\times 170\times {10}^{-3}\times 9.8\\ =3.33N\end{array}$

Now we will find the net torque around the point in which acts. We chose counterclockwise to be the positive torque direction

We found the arm forces to this point, as you see in the figure below.

Note the yellow Right-triangle and you will find that;

The force arm of the upper m g is $R+2Rcos{45}^{°}=R(1+2cos{45}^{°})$

The force arm of $P_{1}is2Rsin{45}^{°}$

And the lower are having the same force arm of R

whereas R is the radius of the balls.

Now we can easily find the net torque which is given by

$\begin{array}{l}\sum {\tau }_{P_3}=0\\ P_3\left(0\right)+[P_1\times 2Rsin{45}^{°}]-[mg\times R(1+2cos{45}^{°})]-\left[mgR\right]+\left[P_{2}R\right]=0\\ 2R{P}_{1}sin{45}^{°}-mgR(1+2cos{45}^{°})-mgR+P_{2}R=0\end{array}$

Noting that R is a common factor. Divide by R;

$2P_{1}sin{45}^{°}-mg(1+2cos{45}^{°})-mg+P_2=0$

Now the only unknown in this equation is , so we will solve for it

$P_1=\frac{mg(1+2cos{45}^{°})+mg-P_2}{2sin{45}^{°}}$

Plug from the given and (2):

$\begin{array}{c}{P}_1=\frac{(170\times {10}^{-3}\times 9.8\times [1+2cos{45}^{°}])+(170\times {10}^{-3}\times 9.8)-3.33}{2sin{45}^{°}}\\ =1.67N\mathrm{...................}\left(3\right)\end{array}$

And hence, from (1) and (3),

$P_3=1.67N$

The magnitude of the force exerted by the left ball on the right ball..

(b)

Now we need to draw the force diagram of the right ball.

Since the ball is in static equilibrium, so

$\sum F_x=0,and\sum F_y=0$

Hence,

$\begin{array}{l}{F}_x-P_1=0\\ F_x=P_1\end{array}$ (4)

And,

$\begin{array}{l}{F}_y-mg=0\\ F_y=mg\end{array}$ (5)

We know that

$F_{LeftonRight}=\sqrt{F_x^2+F_y^2}$

Plug from (4) and (5):

$F_{LeftonRight}=\sqrt{P_1^2+{\left(mg\right)}^2}$

Plug from the given and from (3);

$\begin{array}{c}{F}_{LeftonRight}=\sqrt{1.{67}^2+{(170\times {10}^{-3}\times 9.8)}^2}\\ F_{LeftonRight}=2.36N\end{array}$