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A block of 50 N weight rests in limiting equilibrium on rough inclined plane whose slope is 30°. The plane is raised to slope 60°. What will be the force F (in N) required acting along the plane that will just move the body up? (Answer up to two decimal places)
Correct answer is '57.74'. Can you explain this answer?
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A block of 50 N weight rests in limiting equilibrium on rough incline...
Given data:
- Weight of the block, W = 50 N
- Slope of the inclined plane initially, θ1 = 30°
- Slope of the inclined plane after raising, θ2 = 60°
- Force required to move the block up, F = ?
Analysis:
To calculate the force required to move the block up the inclined plane, we need to consider the forces acting on the block. These forces include:
1. Weight of the block (mg), acting vertically downwards.
2. Normal force (N), acting perpendicular to the inclined plane.
3. Frictional force (f), acting parallel to the inclined plane.
Step 1: Resolving Forces:
We can resolve the weight of the block into two components:
- The component perpendicular to the inclined plane (mg*cosθ), which is balanced by the normal force (N).
- The component parallel to the inclined plane (mg*sinθ), which is balanced by the frictional force (f).
Step 2: Equating Forces:
In limiting equilibrium, the frictional force (f) is given by the equation:
f = μN
where μ is the coefficient of friction.
Since the block is just about to move up the inclined plane, the frictional force (f) will be at its maximum value, which is given by:
f(max) = μ(max) * N
where μ(max) is the maximum coefficient of friction.
Step 3: Determining the Maximum Coefficient of Friction:
The maximum coefficient of friction is given by:
μ(max) = tan(θ)
where θ is the angle of the inclined plane.
Step 4: Calculating the Force Required:
Substituting the value of μ(max) into the equation for the frictional force (f(max)) gives:
f(max) = μ(max) * N
f(max) = tan(θ) * N
Since the frictional force (f) is equal to the force required to move the block up (F), we have:
F = f(max) = tan(θ) * N
Step 5: Calculation:
Substituting the given values of θ and N into the equation for the force required (F), we get:
F = tan(60°) * N
F = tan(60°) * 50 N
F ≈ 57.74 N
Therefore, the force required to move the block up the inclined plane is approximately 57.74 N.
- Weight of the block, W = 50 N
- Slope of the inclined plane initially, θ1 = 30°
- Slope of the inclined plane after raising, θ2 = 60°
- Force required to move the block up, F = ?
Analysis:
To calculate the force required to move the block up the inclined plane, we need to consider the forces acting on the block. These forces include:
1. Weight of the block (mg), acting vertically downwards.
2. Normal force (N), acting perpendicular to the inclined plane.
3. Frictional force (f), acting parallel to the inclined plane.
Step 1: Resolving Forces:
We can resolve the weight of the block into two components:
- The component perpendicular to the inclined plane (mg*cosθ), which is balanced by the normal force (N).
- The component parallel to the inclined plane (mg*sinθ), which is balanced by the frictional force (f).
Step 2: Equating Forces:
In limiting equilibrium, the frictional force (f) is given by the equation:
f = μN
where μ is the coefficient of friction.
Since the block is just about to move up the inclined plane, the frictional force (f) will be at its maximum value, which is given by:
f(max) = μ(max) * N
where μ(max) is the maximum coefficient of friction.
Step 3: Determining the Maximum Coefficient of Friction:
The maximum coefficient of friction is given by:
μ(max) = tan(θ)
where θ is the angle of the inclined plane.
Step 4: Calculating the Force Required:
Substituting the value of μ(max) into the equation for the frictional force (f(max)) gives:
f(max) = μ(max) * N
f(max) = tan(θ) * N
Since the frictional force (f) is equal to the force required to move the block up (F), we have:
F = f(max) = tan(θ) * N
Step 5: Calculation:
Substituting the given values of θ and N into the equation for the force required (F), we get:
F = tan(60°) * N
F = tan(60°) * 50 N
F ≈ 57.74 N
Therefore, the force required to move the block up the inclined plane is approximately 57.74 N.
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Community Answer
A block of 50 N weight rests in limiting equilibrium on rough incline...
R = W cos 60° ...(1)
F = W sin 60° + μR ...(2)
From equation (1) and (2) we get
= 57.74 N
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A block of 50 N weight rests in limiting equilibrium on rough inclined plane whose slope is 30°. The plane is raised to slope 60°. What will be the force F (in N) required acting along the plane that will just move the body up? (Answer up to two decimal places)Correct answer is '57.74'. Can you explain this answer?
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A block of 50 N weight rests in limiting equilibrium on rough inclined plane whose slope is 30°. The plane is raised to slope 60°. What will be the force F (in N) required acting along the plane that will just move the body up? (Answer up to two decimal places)Correct answer is '57.74'. Can you explain this answer? for Mechanical Engineering 2024 is part of Mechanical Engineering preparation. The Question and answers have been prepared according to the Mechanical Engineering exam syllabus. Information about A block of 50 N weight rests in limiting equilibrium on rough inclined plane whose slope is 30°. The plane is raised to slope 60°. What will be the force F (in N) required acting along the plane that will just move the body up? (Answer up to two decimal places)Correct answer is '57.74'. Can you explain this answer? covers all topics & solutions for Mechanical Engineering 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for A block of 50 N weight rests in limiting equilibrium on rough inclined plane whose slope is 30°. The plane is raised to slope 60°. What will be the force F (in N) required acting along the plane that will just move the body up? (Answer up to two decimal places)Correct answer is '57.74'. Can you explain this answer?.
A block of 50 N weight rests in limiting equilibrium on rough inclined plane whose slope is 30°. The plane is raised to slope 60°. What will be the force F (in N) required acting along the plane that will just move the body up? (Answer up to two decimal places)Correct answer is '57.74'. Can you explain this answer? for Mechanical Engineering 2024 is part of Mechanical Engineering preparation. The Question and answers have been prepared according to the Mechanical Engineering exam syllabus. Information about A block of 50 N weight rests in limiting equilibrium on rough inclined plane whose slope is 30°. The plane is raised to slope 60°. What will be the force F (in N) required acting along the plane that will just move the body up? (Answer up to two decimal places)Correct answer is '57.74'. Can you explain this answer? covers all topics & solutions for Mechanical Engineering 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for A block of 50 N weight rests in limiting equilibrium on rough inclined plane whose slope is 30°. The plane is raised to slope 60°. What will be the force F (in N) required acting along the plane that will just move the body up? (Answer up to two decimal places)Correct answer is '57.74'. Can you explain this answer?.
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A block of 50 N weight rests in limiting equilibrium on rough inclined plane whose slope is 30°. The plane is raised to slope 60°. What will be the force F (in N) required acting along the plane that will just move the body up? (Answer up to two decimal places)Correct answer is '57.74'. Can you explain this answer?, a detailed solution for A block of 50 N weight rests in limiting equilibrium on rough inclined plane whose slope is 30°. The plane is raised to slope 60°. What will be the force F (in N) required acting along the plane that will just move the body up? (Answer up to two decimal places)Correct answer is '57.74'. Can you explain this answer? has been provided alongside types of A block of 50 N weight rests in limiting equilibrium on rough inclined plane whose slope is 30°. The plane is raised to slope 60°. What will be the force F (in N) required acting along the plane that will just move the body up? (Answer up to two decimal places)Correct answer is '57.74'. Can you explain this answer? theory, EduRev gives you an
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