Question

A vendor at a fair ties her tent to a rock-filled coffee can in order to hold the rope down. The can, including the rocks, has a mass of 2.5 kg. A stiff breeze pushes against the tent and cases the rope attached to the can to become taut. The rope makes an angle of 50° with the horizontal and has a tension of 18 N. If the can remains in place, what is the minimum coefficient of static friction between the can and the ground?

Answer 1

Answer: FN=mg-Tsin1-cos=/UNN=3-9.00-Tsin(40)-lTcos140)=Mmg-pitsin0Tcos(401+1.52)sin(405--5213119.00)T=Bl

Answer 2

Answer:

Approximately $1.08$ (assuming that the ground is level.)

Explanation:

The coefficient of static friction between two surfaces is the ratio between:

• Maximum possible magnitude of static friction between the two surfaces, and
• Normal force between the two surfaces.

Both the static friction and the normal force need to be found.

Assuming that the ground in this question is level, forces in the horizontal component would consist of:

• static friction, and
• horizontal component of tension.

Forces in the vertical component would consist of:

• vertical component of tension,
• normal force, and
• weight.

Hence, the magnitude of static friction can be found from the horizontal component of tension. The normal force would be equal to the difference between weight and the vertical component of tension.

Assume that the ground in this question is level. Aside from static friction, the only force in the horizontal direction would come from the rope. At $50^{\circ}$ above the horizon, the $18\; {\rm N}$-tension in the rope consists of two components:

• Horizontal component: $(18\; {\rm N})\, \cos(50^{\circ})$.
• Vertical component: $(18\; {\rm N})\, \sin(50^{\circ})$ (upward.)

In particular, static friction should balance the horizontal component of this tension. Hence, the two forces should be equal in magnitude. In other words, the magnitude of static friction should be $(18\; {\rm N})\, \cos(50^{\circ}) \approx 11.6\; {\rm N}$.

Subtract the (upward) vertical component of tension from weight to find the magnitude of normal force:

$\begin{aligned}(2.5\; {\rm kg})\, (9.81\; {\rm N\cdot kg^{-1})) - (18\; {\rm N})\, \sin(50^{\circ}) \approx 10.7\; {\rm N}\end{aligned}$.

Divide the magnitude of static friction by the magnitude of normal force to find the minimum coefficient of static friction between the two surfaces:

$\begin{aligned}\frac{(\text{static friction})}{(\text{normal force})} &= \frac{11.6\; {\rm N}}{10.7\; {\rm N}} \approx 1.08\end{aligned}$.