Question

A small box of mass 4kg slides down a frictional incline built atop a wedge of mass 5kg. The incline is at an angle of 45°, and the coefficient of friction between the box and the wedge is $\mu_1=0.5$. The wedge sits on a rough horizontal surface with an unknown coefficient of friction $\mu_2$. What is the minimum value of $\mu_2$ required to keep the wedge from sliding, while the box moves down the incline ?

• A. 0.30
• B. 0.125
• C. 0.45
• D. 0.20

Answer 1

The correct answer is 0.375, which is not among the options provided. There seems to be an error in the options.

Answer 2

To find the minimum value of $\mu_2$ required to keep the wedge from sliding, we need to analyze the forces acting on both the box ($m_1$) and the wedge ($m_2$). The wedge is in equilibrium, meaning the net force on it is zero.

Given:

• Mass of box, $m_1 = 4 \, \text{kg}$
• Mass of wedge, $m_2 = 5 \, \text{kg}$
• Angle of incline, $\theta = 45^\circ$
• Coefficient of friction between box and wedge, $\mu_1 = 0.5$
• Acceleration due to gravity, $g = 10 \, \text{m/s}^2$

Step 1: Forces on the Box ($m_1$)

• Normal force ($N_1$): $N_1 = m_1g \cos\theta = 4 \times 10 \times \cos 45^\circ = 20\sqrt{2} \, \text{N}$
• Frictional force ($f_1$): $f_1 = \mu_1 N_1 = 0.5 \times 20\sqrt{2} = 10\sqrt{2} \, \text{N}$

Step 2: Forces on the Wedge ($m_2$)

The wedge is stationary, so the net force on it is zero. The forces exerted by the box ($m_1$) on the wedge ($m_2$) are:

• Normal force $N_1'$: Equal in magnitude to $N_1$ but acts perpendicular to the incline, into the wedge.
• Frictional force $f_1'$: Equal in magnitude to $f_1$ but acts parallel to the incline, downwards along the incline.

Resolve $N_1'$ and $f_1'$ into horizontal and vertical components:

• Horizontal component (to the right): $N_1 \sin 45^\circ = 20\sqrt{2} \times \frac{1}{\sqrt{2}} = 20 \, \text{N}$ and $f_1 \cos 45^\circ = 10\sqrt{2} \times \frac{1}{\sqrt{2}} = 10 \, \text{N}$
• Vertical component (downwards): $N_1 \cos 45^\circ = 20\sqrt{2} \times \frac{1}{\sqrt{2}} = 20 \, \text{N}$ and $f_1 \sin 45^\circ = 10\sqrt{2} \times \frac{1}{\sqrt{2}} = 10 \, \text{N}$

Apply equilibrium conditions for the wedge:

1. Vertical equilibrium: $N_2 = m_2g + N_1 \cos 45^\circ + f_1 \sin 45^\circ = (5 \times 10) + 20 + 10 = 80 \, \text{N}$
2. Horizontal equilibrium: $f_2 = N_1 \sin 45^\circ + f_1 \cos 45^\circ = 20 + 10 = 30 \, \text{N}$

Step 3: Calculate the minimum $\mu_2$

For the wedge to be on the verge of sliding: $\mu_2 = \frac{f_2}{N_2} = \frac{30}{80} = 0.375$

The calculated value of $\mu_2 = 0.375$ is not among the options.