Physics Solution: Rope in Equilibrium
Minimum coefficient of friction required when the external force is applied tangentially to the rope.
Diagram
The applied force \(F\) is shown along the tangent to the rope at the free end.
1. Given Data and Basic Idea
The rope has:
- Total mass \(m = 3\,\text{kg}\)
- Total length \(L\)
- Portion on the ground \(= \eta L\)
- Portion in air \(= (1-\eta)L\)
- Applied force \(F = 10\sqrt{2}\,\text{N}\)
Hence,
$$m_{\text{ground}}=\eta m, \qquad m_{\text{air}}=(1-\eta)m$$
For equilibrium, both horizontal and vertical force balances must hold.
2. Vertical Equilibrium
The portion of rope in air is supported by the vertical component of the applied force.
$$F_y = m_{\text{air}}g = (1-\eta)mg$$
So the vertical component of the external pull must balance the weight of the hanging portion.
3. Horizontal Equilibrium
The horizontal component of the applied force is balanced by friction on the portion of rope lying on the ground.
$$F_x = f$$
Also, from the force triangle,
$$F_x = \sqrt{F^2 – F_y^2}$$
The maximum available friction is:
$$f_{\max} = \mu N$$
For the portion on the ground, the normal reaction is
$$N = m_{\text{ground}}g = \eta mg$$
Therefore, for equilibrium,
$$F_x \le \mu \eta mg$$
4. Condition on \(\mu\)
Substitute \(F_x = \sqrt{F^2 – F_y^2}\) and \(F_y = (1-\eta)mg\):
$$\sqrt{F^2 – \left[(1-\eta)mg\right]^2} \le \mu \eta mg$$
So,
$$\mu \ge \frac{1}{\eta mg}\sqrt{F^2-(1-\eta)^2m^2g^2}$$
Or, more neatly,
$$\mu \ge \frac{1}{\eta}\sqrt{\left(\frac{F}{mg}\right)^2-(1-\eta)^2}$$
5. Numerical Calculation
Given:
$$m=3,\qquad g=10 \quad \Rightarrow \quad mg=30$$
$$F=10\sqrt{2},\qquad \eta=\frac{2}{3}$$
Now,
$$\frac{F}{mg}=\frac{10\sqrt{2}}{30}=\frac{\sqrt{2}}{3}$$
$$\left(\frac{F}{mg}\right)^2=\frac{2}{9}$$
$$(1-\eta)=1-\frac{2}{3}=\frac{1}{3} \quad \Rightarrow \quad (1-\eta)^2=\frac{1}{9}$$
Substitute in the formula:
$$\mu \ge \frac{1}{2/3}\sqrt{\frac{2}{9}-\frac{1}{9}}$$
$$\mu \ge \frac{3}{2}\sqrt{\frac{1}{9}}$$
$$\mu \ge \frac{3}{2}\cdot\frac{1}{3}=\frac{1}{2}$$
Final Answer: \(\boxed{\mu_{\min}=0.5}\)