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 3 of 5.
Ayomidol said:
7 years ago
Answer is A. The horizontal component of the weight.
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?
W tanθ. Is for moving body upward on an inclined plane by neglecting the friction?
Anomi said:
7 years ago
Here it is a miskate in symbol, instead of θ it must be φ.
P=uR=tan φW.
P=uR=tan φW.
Rohit Koundal said:
7 years ago
Answer will be sinθ.
Samir said:
8 years ago
You are absolutely right. @Gaurav.
Deepak said:
8 years ago
If force is applied horizontally and 'θ' is the angle of friction, W tanθ is the right answer.
LOKESH D said:
8 years ago
Option b is correct because in horizontal cos vertical sin and resultant is tan.
Valak said:
8 years ago
Option A is correct.
Jayprakash said:
8 years ago
W sin θ is the right answer. Because of min force required.
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).
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