A box of mass 3 units rests on a horizontal rough surface and is tied to

the end of a rope. Peter pulls on the rope and exerts a force of magnitude sqrt(2g)

with the rope making an angle of �Pi/4 with the horizontal. Given that the box
is on the point of slipping, show that the coeff�cient of friction is 1/2.

To show that the coefficient of friction is 1/2, we need to analyze the forces acting on the box and determine the condition for it to be on the point of slipping. Let's break down the problem step by step:

1. Draw a free-body diagram:
- The weight of the box acts vertically downwards (mg).
- The normal force from the rough surface acts perpendicular to the surface.
- The tension in the rope acts at an angle of π/4 with the horizontal.
- The frictional force opposing the motion acts parallel to the surface and opposite to the direction of motion.

2. Analyze the forces:
The weight of the box can be calculated as mg, where m is the mass of the box and g is the acceleration due to gravity.
The tension in the rope can be calculated as T = √(2g), where g is the acceleration due to gravity.
The normal force is equal to the weight of the box since the box is on a horizontal surface and not accelerating vertically.
The frictional force can be calculated as μN, where μ is the coefficient of friction and N is the normal force.

3. Resolve the forces:
Resolve the tension in the rope into its horizontal and vertical components: Tsin(π/4) and Tcos(π/4).
The vertical component of the tension (Tcos(π/4)) cancels out with the weight of the box (mg), as the box is not accelerating vertically.
The horizontal component of the tension (Tsin(π/4)) is opposed by the frictional force (μN).
Equate these two forces to determine the coefficient of friction:
Tsin(π/4) = μN
√(2g) * sin(π/4) = μ * mg

4. Simplify the equation:
Using the identity sin(π/4) = cos(π/4) = 1/√2, we can simplify the equation:
√(2g) * 1/√2 = μ * mg
√g = μg

5. Solve for the coefficient of friction:
Divide both sides of the equation by g to isolate μ:
√g / g = μ
μ = 1/√g

6. Substitute the value of g:
μ = 1/√(9.8) = 1/√(2 * 4.9) = 1/√2 * 1/√4.9 ≈ 1/2

Therefore, the coefficient of friction (μ) is approximately 1/2, showing that the box is on the point of slipping.