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 5 of 5.
Anand Sharma said:
8 years ago
In horizontal, plane factor will be Wcosθ.
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)
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).
Where u=coefficient of friction which is equal to tan(θ)
N= normal reaction which is equal to W,
therefore P= W tan(θ).
Jayprakash said:
8 years ago
W sin θ is the right answer. Because of min force required.
Valak said:
8 years ago
Option A is correct.
LOKESH D said:
8 years ago
Option b is correct because in horizontal cos vertical sin and resultant is tan.
Deepak said:
8 years ago
If force is applied horizontally and 'θ' is the angle of friction, W tanθ is the right answer.
Andile said:
1 decade ago
How come because its a horizontal plane?
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