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A vertical dock gate 2 meter wide remains in position due to horizontal force of water on one side. The gate weights 800 Kg and just starts sliding down when the depth of water upto the bottom of the gate decreases to 4 meters. Then the coefficient of friction between dock gate and dock wall will be:  
  • a)
    0.5
  • b)
    0.2
  • c)
    0.05
  • d)
    0.02
Correct answer is option 'C'. Can you explain this answer?
Verified Answer
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
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Most Upvoted Answer
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.
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Community Answer
A vertical dock gate 2 meter wide remains in position due to horizonta...
C
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A vertical dock gate 2 meter wide remains in position due to horizontal force of water on one side. The gate weights 800 Kg and just starts sliding down when the depth of water upto the bottom of the gate decreases to 4 meters. Then the coefficient of friction between dock gate and dock wall will be:a)0.5b)0.2c)0.05d)0.02Correct answer is option 'C'. Can you explain this answer? for Mechanical Engineering 2025 is part of Mechanical Engineering preparation. The Question and answers have been prepared according to the Mechanical Engineering exam syllabus. Information about A vertical dock gate 2 meter wide remains in position due to horizontal force of water on one side. The gate weights 800 Kg and just starts sliding down when the depth of water upto the bottom of the gate decreases to 4 meters. Then the coefficient of friction between dock gate and dock wall will be:a)0.5b)0.2c)0.05d)0.02Correct answer is option 'C'. Can you explain this answer? covers all topics & solutions for Mechanical Engineering 2025 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for A vertical dock gate 2 meter wide remains in position due to horizontal force of water on one side. The gate weights 800 Kg and just starts sliding down when the depth of water upto the bottom of the gate decreases to 4 meters. Then the coefficient of friction between dock gate and dock wall will be:a)0.5b)0.2c)0.05d)0.02Correct answer is option 'C'. Can you explain this answer?.
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