Mechanical Engineering - Engineering Mechanics - Discussion

Discussion Forum : Engineering Mechanics - Section 2 (Q.No. 7)

Engineering Mechanics

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.

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)

Somesh said: 6 years ago

Answer C is correct, because it is a case of body slide down an inclined plane. Incase of the horizontal answer, A is correct.

Amit said: 1 decade ago

Weight w acts downward so that to move weight w we should only give force equals to frictional force. It will be wtanthita.

Muniraju B said: 7 years ago

The answer is wsinθ

W tanθ. Is for moving body upward on an inclined plane by neglecting the friction?

Phoring said: 6 years ago

Option A is the right answer. wsin(θ) is the horizontal component which makes the body to move or slide.

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

Deepak said: 8 years ago

If force is applied horizontally and 'θ' is the angle of friction, W tanθ is the right answer.

Anomi said: 7 years ago

Here it is a miskate in symbol, instead of θ it must be φ.

P=uR=tan φW.

Dhananjay said: 4 years ago

@Sandaz.

But we don't if the force applied is horizontal or at an angle to the surface.

LOKESH D said: 8 years ago

Option b is correct because in horizontal cos vertical sin and resultant is tan.

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