Mechanical Engineering - Engineering Mechanics - Discussion
Discussion Forum : Engineering Mechanics - Section 2 (Q.No. 7)
7.
The minimum force required to slide a body of weight W on a rough horizontal plane is
Discussion:
49 comments Page 1 of 5.
Yugal said:
6 years ago
The answer is A.
The minimum force required to slide a body of weight W on a rough horizontal plane is W sinϴ. A body of weight W is required to move up the rough inclined plane whose angle of inclination with the horizontal is α. The effort applied parallel to the plane is given by P = W (sinα + μ cosα).
The minimum force required to slide a body of weight W on a rough horizontal plane is W sinϴ. A body of weight W is required to move up the rough inclined plane whose angle of inclination with the horizontal is α. The effort applied parallel to the plane is given by P = W (sinα + μ cosα).
Akash said:
3 years ago
@All.
Actually, the question is not properly defined.
When a horizontal force is applied to the block the W tan (theta) is the right answer.
But when we apply a force that is inclined to the horizontal instead of a horizontal force then the minimum force is Wsin (θ).
Actually, the question is not properly defined.
When a horizontal force is applied to the block the W tan (theta) is the right answer.
But when we apply a force that is inclined to the horizontal instead of a horizontal force then the minimum force is Wsin (θ).
(5)
Gaurav1995 said:
9 years ago
Given ans is absolutely correct.
The body lies on a horizontal plane. So to slide the body force has to overcome a limiting frictional force of the body.
i.e. F = uW where, u = coeff of friction = Tanθ (θ = friction angle).
The body lies on a horizontal plane. So to slide the body force has to overcome a limiting frictional force of the body.
i.e. F = uW where, u = coeff of friction = Tanθ (θ = friction angle).
Sandaz said:
4 years ago
Yes, the Correct answer is tan(θ). Those who are saying cosθ' and sinθ, that is for inclined plane and not for horizontal surface. Here in question, it is only horizontal surface, so it's option C.
Kanchumarthi Naga Sai Durga said:
8 years ago
The weight acts on a rough horizontal plane so answer is W cos(θ). If the force acts on a rough vertical plane answer is W sin(theta).
Horizontal plane - W cos(θ).
Vertical plane - W sin(θ).
Horizontal plane - W cos(θ).
Vertical plane - W sin(θ).
Parth said:
9 years ago
Right answer: W sin theta because in question already specified that minimum force. So whenever angel of inclination = angel of friction for rough horizontal plane then we have to put the answer W sin theta.
Mohammed Nayeem said:
8 years ago
In the case of a horizontal plane, P = u * N.
Where u=coefficient of friction which is equal to tan(θ)
N= normal reaction which is equal to W,
therefore P= W tan(θ).
Where u=coefficient of friction which is equal to tan(θ)
N= normal reaction which is equal to W,
therefore P= W tan(θ).
Dhananjay said:
4 years ago
If theta is the angle at which force is applied then the answer is A.
And If theta is friction angle and force is applied parallel to base then the answer is C.
And If theta is friction angle and force is applied parallel to base then the answer is C.
(1)
Rajat said:
7 years ago
If we will apply a force at an angle equal to the angle of friction. Minimum force is required to slide that is equal to wsinθ. A is correct.
Roni said:
1 decade ago
F equals to uR and u equals to tan of angle of friction. And R equals to W for horizontal plane. On this perspective, it should be the option (C).
Post your comments here:
Quick links
Quantitative Aptitude
Verbal (English)
Reasoning
Programming
Interview
Placement Papers