How to account for air resistance in a sphere being released against a flow or air?

Jun 2017
337
6
Lima, Peru
The problem is as follows:

A bob is hanging from the ceiling of a specially designed room where a flow of air is being blown against. The bob is makes a $53^{\circ}$ angle with respect of the ground as indicated in the figure. The mass of the sphere is $2\,kg$. Assume that the wire is ideal. The flow of air excerts a constant force whose modulus is $4\,N$. Given these conditions, find the modulus of the force in $N$ in the wire when the sphere passes through its lowest point.



The alternatives are as follows:

$\begin{array}{ll}
1.&26.9\,N\\
2.&29.6\,\frac{m}{s^2}\\
3.&27.5\,\frac{m}{s^2}\\
4.&23.3\,\frac{m}{s^2}\\
5.&21.2\,\frac{m}{s^2}\\
\end{array}$

I'm not sure exactly how should I account for the force of air going against the bob.

What I think should be used here is the conservation of mechanical energy:

$E_u=E_k$

$mgh=\frac{1}{2}mv^2$

When the ball passes through the lowest point will be:

$T-mg=\frac{mv^2}{R}$

$T=\frac{mv^2}{R}+mg$

From the first equation:

$mgh=\frac{1}{2}mv^2$

$2g(1-\sin 53^{\circ})=\frac{v^2}{R}$

Therefore the tension will be:

$T=mg+\frac{mv^2}{R}=mg+2mg(1-\sin 53^{\circ})$

Therefore:

$T=2\times 10 + 2 \times 10 (1-\frac{4}{5})= 20 +20(\frac{1}{5})=24\,N.$ But this is not within the alternatives, needless to say that this doesn't seem to be the right answer. Can someone help here please?. I'm still stuck here.
 

skeeter

Math Team
Jul 2011
3,266
1,766
Texas
the 53 degree angle in the diagram is with respect to the vertical, not the ground (the horizontal), so I don't know which is correct, the diagram or the statement.

let $L$ be the length of the wire and $\theta$ be the angle the wire makes with the vertical.

using energy ...

$mgL(1-\cos{\theta}) - 4L\sin{\theta} = \dfrac{1}{2}mv^2$ , where $v$ is the speed at the bottom of the arc

using centripetal force at the bottom of the arc ...

$T - mg = \dfrac{mv^2}{L}$

you now have the necessary equations to determine the tension in the wire
 
Jun 2017
337
6
Lima, Peru
the 53 degree angle in the diagram is with respect to the vertical, not the ground (the horizontal), so I don't know which is correct, the diagram or the statement.

let $L$ be the length of the wire and $\theta$ be the angle the wire makes with the vertical.

using energy ...

$mgL(1-\cos{\theta}) - 4L\sin{\theta} = \dfrac{1}{2}mv^2$ , where $v$ is the speed at the bottom of the arc

using centripetal force at the bottom of the arc ...

$T - mg = \dfrac{mv^2}{L}$

you now have the necessary equations to determine the tension in the wire
The angle $53^{\circ}$ was referring with respect to the vertical. I had a mistake in the transcribing the problem.

But the source of my confusion is where did you arrived to $4L\sin{\theta}$ can you explain this part? The part where you say the angle $\theta$ is with respect to the vertical is confusing. Can you add a FBD for this?. What sort of energy is this? Because im seeing it more as of a force but not an energy in the equation of conservation of mechanical energy.
 

skeeter

Math Team
Jul 2011
3,266
1,766
Texas
But the source of my confusion is where did you arrived to 4Lsinθ can you explain this part?
Work done against the motion of the bob in the horizontal direction ...

$W = F \cdot \Delta x = 4 \cdot L\sin{\theta}$

The part where you say the angle θ is with respect to the vertical is confusing.
In your diagram, the angle labeled 53 degrees is w/ respect to the vertical. I changed it to $\theta$ because I was uncertain that it was 53 degrees w/respect to the vertical ... you originally stated it was 53 degrees w/respect to the ground.