Question

A couple short questions. Thanks! Continuation of my previous question

10. What is the ratio of (mglmo)? Discuss from this expression whether pulling at an angle is ever better than pulling horizontally. Does the coefficient of kinetic friction in the system matter? (What if there is no friction? What if is large?) 11. Based on your expression above, what angle would you recommend pulling the sled with your tractor (given a coefficient of kinetic friction pu)? [Hint: derivatives are useful for maximization problems. Also, two useful derivative formulas are: d(sin0)/dθ cos, and

Answer 1

Consider a mass M on a rough surface being pulled by a force F with a constant acceleration at an angle as shown in the figure, and its free-body diagram is also shown.

theta *g

Applying Newton's second law to the system in vertical and horizontal direction gives:

$N+F*\sin \theta =m*g$ in vertical direction and $F*\cos \theta -\mu _{k}*N=m*a$ in horizontal

Solving for a gives , $F*\cos \theta -\mu _{k}*(m*g-F*\sin \theta )=m*a$

$\therefore a=\frac{F*(\cos \theta +\mu _{k}*\sin \theta)}{m}-\mu _{k}*g$

To find the maximum value of acceleration a, differentiating the expression by theta and equating to zero gives,

$\frac{\mathrm{d} a}{\mathrm{d} t}=\frac{F*(-\sin \theta +\mu _{k}*\cos \theta)}{m}=0$

$\therefore \mu _{k}=\tan \theta$

Thus, we get maximum acceleration of the mass being pulled with a constant force at the angle given by above expression.

10. Pulling at an angle can be better than pulling horizontally when the angle is equal to or close to the angle $\theta =\tan^{-1}\mu _{k}$ . This angle is purely dependent on the coefficient of kinetic friction.If their is no friction then angle would be 0, and if friction is too large then the angle would be greater than 45 degrees.

11. Based on this expression the tractor should pull the sled at an angle given by $\theta =\tan^{-1}\mu _{k}$ .