What Is the Minimal Tension Angle for a Mass Pulled Horizontally?

[Bruneleski]

Homework Statement

Body of mass m is being pulled by a string so that it moves horizontally at constant speed.
Friction coefficient between surface and body is u.Find an angle (alpha) between a string and horizontal surface for which tension in string is minimal and find that tension.

Homework Equations

weight w=mg ; n=normal force
friction force=f=un
acceleration due x-axis=0
tension T

The Attempt at a Solution

<br /> \sum F_y=Tsin(\alpha)+n+(-w)=0; \\<br /> T=\frac{(mg-n)}{sin(\alpha)} \\<br /> \sum F_x=Tcos(\alpha)+(-f)=0 \\<br /> \sum F_x=Tcos(\alpha)+(-un)=0; \: Tcos\alpha=un ; \\<br /> T=\frac{umg}{cos\alpha} \\<br /> \frac{(mg-n)}{sin(\alpha)}=\frac{umg}{cos\alpha} \\<br /> tan\alpha=\frac{mg-n}{umg}<br />

Now I'm not sure what to do next.I need to find minimal angle so derivative of the last expression should be zerr should I differentiate arctan?I'm confused.[/B]

 

[rude man]

I can't follow what you wrote. What are f and n?

You might consider: express F_x as a function of F_y and constants where T² = F_x² + F_y².

Then, express F_x and F_y as functions of constants and θ.
Then find minimum of T in the usual way.
Warning: the math is a bit messy but the answer is beautifully simple.

 

[mfb]

Now I'm not sure what to do next.I need to find minimal angle so derivative of the last expression should be zero.Or should I differentiate arctan?I'm confused.

Don't cancel the tension if you are interested in it. You are not interested in the normal force, so you can get rid of this, that gives you an expression with tension and angle in it.

 

[Bruneleski]

@rude man , Note what i wrote under 2. : f - friction force, n - normal force
@mfb I'm not sure how to connect these

T=umg/cosα
T=(mg-n)/sinα

If i just differentiate T=umgcosx, I get nonsensical angle
<br /> T&#039;=-\frac{umg}{cos^2(\alpha)}<br />

So this to be minimum(zero) , angle should be 90 degrees which makes no sense

 

[mfb]

The normal force is not mg, you have to consider this for the horizontal part (you did that correctly for the vertical part).

I'm not sure how to connect these

You have two equations with three unknown parameters - tension, angle, and normal force. This does not have a unique solution. To find the angle of minimal tension, it would be nice to have an equation that depends on tension and angle only (there you can follow the standard approach with the derivative), but not on the unknown normal force any more. You can solve one equation for the normal force and plug it into the other to get that.