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.
Ayyavu said:
10 years ago
W cos is answer. Force act in a horizontal.
Rajesh khutdar said:
9 years ago
we know F = w R,
w = tan theta and for horizontal surface R = W
Therefore, Force required (F) = w tan theta
w = tan theta and for horizontal surface R = W
Therefore, Force required (F) = w tan theta
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.
Parth pandya said:
9 years ago
I think, Option A is right answer.
Abrha said:
9 years ago
The correct answer is option A.
The mechanic said:
9 years ago
@Rajesh
You are right, I agree with you.
You are right, I agree with you.
Dalveer said:
9 years ago
Why not A? Please explain it.
Rohit Koundal said:
7 years ago
Answer will be sinθ.
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).
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).
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