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
W sin θ
W cos θ
W tan θ
none of these
Answer: Option
Explanation:
No answer description is available. Let's discuss.
Discussion:
49 comments Page 2 of 5.

The mechanic said:   9 years ago
@Rajesh

You are right, I agree with you.

Dalveer said:   9 years ago
Why not A? Please explain it.

Nik said:   9 years ago
I think option A is the right answer.

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).

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).

Anand Sharma said:   8 years ago
In horizontal, plane factor will be Wcosθ.

Vinit said:   8 years ago
According to me, the correct answer is A.
(1)

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)

Hareesh said:   8 years ago
I agree with the given answer.

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).


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