Uniform rod AB with end B resting against rough vertical wall. Coefficient of friction between wall and rod is μ.
Rod is 2m long and has mass 3kg.
Rod is kept in limiting equilibrium by a light inextensible string, one end of which is attached to the end A of the rod and the other to point C on the wall 2m above B. Angle ABC=120°.
The end B of the rod is about to slip down the wall.
1. By taking moments about C, show that the normal reaction at B is approx. 12.7N.
2. Find the value of μ.
PS: A labelled diagram would be very helpful! Thanks!
To solve the problem, let's start by drawing a diagram:
```
B
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A C
```
1. To find the normal reaction at B, we need to calculate the sum of moments acting on the rod about point C.
Taking moments about C:
Sum of clockwise moments = Sum of anti-clockwise moments
Clockwise moments:
There are two clockwise moments acting on the rod - the weight of the rod (mg) and the tension in the string (T).
Clockwise moment due to weight = mg * (2/3) * cos(30°) = (3 kg * 9.8 m/s^2 * (2/3) * cos(30°))
Anti-clockwise moments:
The only anti-clockwise moment acting on the rod is the frictional force between the wall and the rod, which is given by μ * Normal reaction (N) at B.
Setting the clockwise moments equal to the anti-clockwise moments, we have:
(3 kg * 9.8 m/s^2 * (2/3) * cos(30°)) = μ * N
Simplifying, we get:
N = (3 kg * 9.8 m/s^2 * (2/3) * cos(30°)) / μ
Substituting the given values, we can calculate the approximate value of N.