Terminal Velocity and Mass Relationship in a Proportionality Argument

[opticaltempest]

Homework Statement

Assume a box is sliding down a ramp with an incline of $$\theta$$ radians and reaches terminal velocity before arriving at the bottom of the ramp. Assume that the drag force caused by the air is proportional to $$Sv^2$$, where $$S$$ is the cross sectional area perpendicular to the direction of motion and $$v$$ is the speed. Further assume that the sliding friction between the object and the ramp is proportional to the normal weight of the object. Determine the relationship between the terminal velocity and the mass of the object.

The Attempt at a Solution

The forces acting on the box will be the drag force $$F_d$$ acting in the negative direction, the sliding friction $$F_s$$ acting in the negative direction, and the component of gravity that is parallel to the surface of the ramp $$F_g_x$$ acting in the positive direction. At terminal velocity, the box is not accelerating. Using Newton's Second Law for the net force acting on the box, we have

$$-F_s-F_d+F_g_x=0$$ (1)

Now, $$F_s \propto w$$ and $$w \propto m$$ so

$$F_s \propto m$$

where $$w$$ is the normal weight of the box and $$m$$ is the mass of the box.

The same argument can be made for $$F_g_x$$. So

$$F_g_x \propto m$$.

For $$F_d$$, note that $$S \propto L^2$$ and $$V \propto L^3$$,

where $$L$$ is length and $$V$$ is volume.

So $$L\propto S^{\frac{1}{2}} \propto V^{\frac{1}{3}}$$.

This implies that $$S \propto V^{\frac{2}{3}}$$.

Since volume $$V$$ is proportional to mass $$m$$, we have

$$S \propto m^{\frac{2}{3}}$$.

Assuming we are at terminal velocity, $$v=v_T$$.

Now going back to (1), we have

$$-F_s-F_d+F_g_x=0$$

$$-m-m^{\frac{2}{3}}v^2_{T}+m \propto 0$$

$$m^{\frac{2}{3}}v^2_{T} \propto 0$$ (2)

The proportionality in (2) doesn't make sense. Where am I going wrong? Do I need to modify

$$F_g_x \propto m$$ to be $$F_g_x \propto \sin{(\theta)}m$$.?Thanks

 

[dynamicsolo]

The forum is giving me a "Page Not Found" when I try to post my reply. I'll try PMing it to you...

...OK, the system doesn't like something about my post in PM as well. Write me and I'll discuss it with you there.

 

[fleem]

If you have an expression with multiple terms:

$$-F_s-F_d+F_g_x$$

You can't replace each term with values proportional to each term:

$$-m-m^{\frac{2}{3}}v^2_{T}+m$$

This is because each term has a different (hidden) constant. Try doing the problem with equalities rather than proportions, remembering to show the constant in each term. I think you'll see the mistake then.

Another thing is that you are making certain assumptions about the problem that could likely be wrong. For example, it doesn't say that the volume is proportional to the mass. Although it is rather unrealistic that the dynamic friction is unrelated to the velocity, but I guess we are to assume that.

 

[opticaltempest]

Fleem, I think I see the point you are making.

Let $$F_s=Km$$,

$$F_g_x=Rm$$, and

$$F_d=Qm^{\frac{2}{3}} v_{t}^{2}$$,

where K, Q, and R are some contants.

Equation (1) from my first post becomes

$$-Km-Qm^{\frac{2}{3}}v^{2}_{t}+Rm=0$$

$$\implies -Qm^{\frac{2}{3}}v^{2}_{t}=m(K-R)$$

$$\implies v^{2}_{t}=\frac{m(K-R)}{m^{\frac{2}{3}}(-Q)}$$

$$\implies v = \sqrt{\frac{K-R}{-Q}}\sqrt{m^{\frac{1}{3}}}$$

$$\implies v = \sqrt{\frac{K-R}{Q}}m^{\frac{1}{6}}$$

Let $$T=\sqrt{\frac{K-R}{Q}}$$. So,

$$v = Tm^{\frac{1}{6}}$$

$$\implies v \propto m^{\frac{1}{6}}$$.

So by replacing each term with what it was proportional to, I was assuming that all the proportionality constants were the same, which they might not be. Right?

I should also say that I am assuming that volume is proportional to mass. I guess a better assumption would be to assume some constant uniform density for all boxes. From that assumption, volume would be proportional to mass. Right? The book makes the assumption that volume is proportional to mass.

 

[fleem]

Just glanced over the algebra but, yep, that's the idea.