Q64P - University Physics with Modern Physics

Question

A horizontal wire holds a solid uniform ball of mass in place on a tilted ramp that rises $\mathbf{35}\mathbf{.}\mathbf{0}\mathbf{°}$ above the horizontal. The surface of this ramp is perfectly smooth, and the wire is directed away from the center of the ball (Fig. P5.64). 

(a) Draw a freebody diagram of the ball. 

(b) How hard does the surface of the ramp push on the ball? 

(c) What is the tension in the wire?

Step-by-Step Solution

Answer

(a) 

(b) The surface of the ramp has to push force of 1.22mg on the wall.

(c) The tension in the wire is 0.7mg.

Step 1: Identification of given data

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 $\mathrm{\theta }=35.0°$ .

Step 2: Significance of the tension

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.

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

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$\theta$ and $mg \cos \theta$ respectively. The ball is also exerting a tension T that also has two components such as  T sin $\theta$ and $T \cos \theta$ respectively. The ground is also exerting a reaction force R on the ball.

Step 4: (b) Determination of the force the surface on the ramp push 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:

 

 $\sum _{ }{F}_x=0$

 

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

 

The above equation can also be expressed as:

 

$R=mg\cos \theta +T\sin \theta$                                                                                                         (i)

 

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

 

$\sum _{ }{F}_y$ = 0 

 

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

 

The above equation can also be expressed as:

 

$$\begin{gathered} \mathrm{Tcos\theta }=\mathrm{mgsin\theta } \\ \mathrm{T}=\mathrm{mgtan\theta } \end{gathered}$$                                                                                                         (ii)

 

Substitute $mg\tan \theta$ for T in the equation (i), and we get,

 

$R=mg\cos \theta +mg\tan \theta \sin \theta$

 

Substitute $35.0°$ for $\theta$ in the above equation, and we get,

 

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

 

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

Step 5: (c) Determination of the tension in the wire

The equation (ii) is recalled again.

 

$$\begin{gathered} \mathrm{Tcos\theta }=\mathrm{mgsin\theta } \\ \mathrm{T}=\mathrm{mgtan\theta } \end{gathered}$$

 

Substitute $35.0°$ for $\theta$ 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 .