A vertical dock gate 2 meter wide remains in position due to horizonta...
Ans. (c)
μP =W
or
μρg(4´ 2).(4 / 2) = 800 x g
or
μ = 0.05
View all questions of this test
A vertical dock gate 2 meter wide remains in position due to horizonta...
Explanation:
To find the coefficient of friction between the dock gate and dock wall, we can use the concept of equilibrium of forces. When the gate just starts sliding down, the force of friction is equal to the horizontal force of water on the gate.
1. Write down the given information:
- Width of the dock gate = 2 meters
- Weight of the dock gate = 800 Kg
- Depth of water at which gate starts sliding down = 4 meters
2. Calculate the horizontal force of water on the gate:
The horizontal force of water on the gate is equal to the pressure of water at the depth of 4 meters multiplied by the area of the gate.
Pressure of water = density of water x gravitational acceleration x depth of water
= 1000 kg/m3 x 9.81 m/s2 x 4 m
= 39240 N/m2
Horizontal force of water on the gate = pressure of water x area of gate
= 39240 N/m2 x 2 m
= 78480 N
3. Calculate the force of friction:
Since the gate just starts sliding down, the force of friction is equal to the horizontal force of water on the gate. Therefore,
Force of friction = 78480 N
4. Calculate the normal force:
The normal force is equal to the weight of the gate, which is 800 Kg x 9.81 m/s2 = 7848 N.
5. Calculate the coefficient of friction:
The coefficient of friction is defined as the ratio of the force of friction to the normal force. Therefore,
Coefficient of friction = force of friction / normal force
= 78480 N / 7848 N
= 10
However, the coefficient of friction cannot exceed the value of 1, as it is a ratio of two forces. Therefore, the correct answer is option C, i.e., 0.05.
A vertical dock gate 2 meter wide remains in position due to horizonta...
C