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 4 of 5.
Hareesh said:
8 years ago
I agree with the given answer.
Jay dewangan said:
8 years ago
P = Wsinfi/cos(θ-fi).
We know that, angle of inclinstion = angle of friction for rough horizontal plane. So that answer is Wsin(θ)
We know that, angle of inclinstion = angle of friction for rough horizontal plane. So that answer is Wsin(θ)
Vinit said:
8 years ago
According to me, the correct answer is A.
(1)
Anand Sharma said:
8 years ago
In horizontal, plane factor will be Wcosθ.
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).
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).
Nik said:
9 years ago
I think option A is the right answer.
Dalveer said:
9 years ago
Why not A? Please explain it.
The mechanic said:
9 years ago
@Rajesh
You are right, I agree with you.
You are right, I agree with you.
Abrha said:
9 years ago
The correct answer is option A.
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