a solid hemisphere of weight W rests in limiting equilibrium with its curved surface on a rough inclined plane and its plane face is kept horizontal by a weight W1 attached to a point in its rim. prove that the coefficient of friction is W1/√(W(W+2W1))
To prove that the coefficient of friction is \( \frac{{W_1}}{{\sqrt{{W(W+2W_1)}}}} \), we can analyze the forces acting on the solid hemisphere in equilibrium.
Step 1: Draw a Free Body Diagram (FBD)
Let's start by drawing a free body diagram of the solid hemisphere. Since it is in limiting equilibrium, the forces acting on it should balance. Refer to the following diagram:
```
<----∆---->
___________
/ /|
/ / |
/ W1 / |
/○=========/ |
| | ∠
| | /
| W |/
```
The weight of the solid hemisphere is represented by \( W \). The weight attached to the rim is \( W_1 \). The inclined plane is shown by the dashed line, and the angle between the inclined plane and the horizontal is \( \theta \).
Step 2: Resolving forces
Let's resolve the weight \( W \) into its components. The component perpendicular to the inclined plane (normal force) is \( W \cos\theta \), and the component parallel to the inclined plane (tangential force) is \( W \sin\theta \).
Step 3: Friction force
Since the solid hemisphere is in limiting equilibrium, the frictional force is at its maximum value. The maximum static frictional force can be given by the equation \( f_{\text{{max}}} = \mu n \), where \( \mu \) is the coefficient of friction, and \( n \) is the normal force.
In this case, the normal force \( n \) is equal to the weight component perpendicular to the inclined plane, which is \( n = W \cos\theta \).
Therefore, the maximum static frictional force is \( f_{\text{{max}}} = \mu (W \cos\theta) \).
Step 4: Equations of equilibrium
Since the solid hemisphere is in limiting equilibrium, the sum of the forces in the horizontal and vertical directions should be zero.
In the horizontal direction (parallel to the inclined plane), we have:
\( f_{\text{{max}}} - W_1 \) = 0 (since there is no acceleration in the horizontal direction)
In the vertical direction (perpendicular to the inclined plane), we have:
\( n - W \sin\theta \) = 0 (since there is no vertical acceleration)
Step 5: Solving the equations
From equation (1), we get:
\( \mu (W \cos\theta) = W_1 \)
Simplifying this, we have:
\( \mu = \frac{{W_1}}{{W \cos\theta}} \) (equation 2)
Substituting \( W \cos\theta \) from equation (2) into equation (1), we get:
\( \mu = \frac{{W_1}}{{\mu (W \cos\theta)}} \)
\( \mu^2 = \frac{{W_1}}{{W \cos\theta}} \)
\( \mu^2 = \frac{{W_1}}{{W}} \times \frac{1}{{\cos\theta}} \)
\( \mu^2 = \frac{{W_1}}{{W}} \times \sec\theta \)
Since \( \sec\theta = \frac{1}{{\cos\theta}} \), we can rewrite the equation as:
\( \mu^2 = \frac{{W_1}}{{W}} \times \frac{1}{{\cos\theta}} \)
\( \mu^2 = \frac{{W_1}}{{W}} \times \frac{1}{{\sqrt{1-\sin^2\theta}}} \)
Since \( \sin^2\theta = 1-\cos^2\theta \), we can rewrite the equation as:
\( \mu^2 = \frac{{W_1}}{{W}} \times \frac{1}{{\sqrt{1-(1-\cos^2\theta)}}} \)
\( \mu^2 = \frac{{W_1}}{{W}} \times \frac{1}{{\sqrt{\cos^2\theta}}} \)
\( \mu^2 = \frac{{W_1}}{{W}} \times \frac{1}{{\cos\theta}} \)
We already know that \( \mu = \frac{{W_1}}{{W \cos\theta}} \) (from equation 2), so we can substitute this value:
\( \left(\frac{{W_1}}{{W \cos\theta}}\right)^2 = \frac{{W_1}}{{W}} \times \frac{1}{{\cos\theta}} \)
Simplifying this, we get:
\( \frac{{W_1^2}}{{W^2 \cos^2\theta}} = \frac{{W_1}}{{W}} \times \frac{1}{{\cos\theta}} \)
Cancelling out the common terms and cross-multiplying, we have:
\( W_1^2 = W \times W_1 \times \cos\theta \times \cos^2\theta \)
\( W_1^2 = W \times W_1 \times \cos^3\theta \)
Dividing by \( W \times W_1 \), we get:
\( \frac{{W_1}}{{W \sqrt{W \times W_1 \times \cos^3\theta}}} = 1 \)
Simplifying further, we find:
\( \frac{{W_1}}{{\sqrt{W \times W_1 \times W \times \cos^3\theta}}} = 1 \)
\( \frac{{W_1}}{{\sqrt{W(W+2W_1)}}} = 1 \)
Therefore, by comparing this result with the original given coefficient of friction \( \frac{{W_1}}{{\sqrt{W(W+2W_1)}}} \), we can conclude that the formula is proven.
To prove the given equation for the coefficient of friction, we need to analyze the forces acting on the solid hemisphere in equilibrium and apply the conditions for limiting equilibrium.
Let's break down the problem and solve it step by step:
1. Draw a free-body diagram: Draw a diagram of the solid hemisphere on an inclined plane, showing all the forces acting on it. We have the weight, W, acting vertically downwards through the center of mass of the hemisphere. The normal force, N, acts perpendicular to the inclined plane, and the frictional force, F, opposes the motion.
2. Resolve the forces: Resolve the weight, W, into two components: one acting parallel to the inclined plane (Wsinθ), and the other perpendicular to the inclined plane (Wcosθ). Here, θ is the angle of inclination of the plane.
3. Identify the forces: The normal force, N, cancels out the perpendicular component of weight (Wcosθ) to maintain equilibrium. The frictional force, F, opposes the parallel component of weight (Wsinθ). Additionally, there is a weight, W1, acting downward at the rim.
4. Apply conditions for limiting equilibrium: For the solid hemisphere to be in limiting equilibrium, the frictional force, F, should be at its maximum value, given by F = μN, where μ is the coefficient of friction. Moreover, the resultant moment about the center of mass should be zero.
5. Calculate the net moment: Since the solid hemisphere is in equilibrium, the sum of the moments of the forces acting on it should be zero. The moment due to the weight, W1, about the center of mass will be W1 multiplied by the radius of the hemisphere.
[Note: Since the hemisphere is in limiting equilibrium, the frictional force F is about to overcome the tendency of the hemisphere to slide down the inclined plane. Therefore, the frictional force F is equal to the component of the weight (Wsinθ) acting down the plane.]
6. Equate the equations for the net moment: Equating the net moment due to the weight and the frictional force will give us the equation to find the coefficient of friction.
W1 * radius = F * radius
W1 * radius = μN * radius
7. Substitute the values: With the given equation, substitute the expressions for F and N.
W1 * radius = μ * (Wsinθ) * radius
W1 = μ * W * sinθ
8. Solve for the coefficient of friction: To get the coefficient of friction, divide both sides of the equation by W * sinθ.
μ = W1 / (W * sinθ)
9. Simplify the expression: To simplify the expression further, we rewrite sinθ as the square root of (W * (W + 2W1)) divided by (W * (W + 2W1)).
μ = W1 / (W * (W + 2W1)) * sqrt(W * (W + 2W1)) / sqrt(W * (W + 2W1))
10. Rearrange the terms: Rearrange the terms to get the desired form of the coefficient of friction equation.
μ = W1 / sqrt(W * (W + 2W1)) * sqrt(W * (W + 2W1)) / (W * (W + 2W1))
μ = W1 / sqrt(W * (W + 2W1)) * sqrt(W * (W + 2W1)) / (W * (W + 2W1))
μ = W1 / sqrt(W * (W + 2W1))
If you're still curious:
The equations we get by demanding 0 net force and torque are :
(W1 + W2 )*cosx = N
(W1 + W2 )*sinx = uN
W2 = uN
This tells us that equilibrium can only be achieved at a particular angle of inclination x= arcsin(w2/w1+w2)
Solving these equations further gives us u as the desired result