Mechanical Engineering - Engineering Mechanics - Discussion
Discussion Forum : Engineering Mechanics - Section 1 (Q.No. 9)
9.
A body of weight W is required to move up on rough inclined plane whose angle of inclination with the horizontal is α. The effort applied parallel to the plane is given by(where μ = tanφ = Coefficient of friction between the plane and the body.)
Discussion:
21 comments Page 1 of 3.
Uttada saikumar said:
6 years ago
P=W sinα + F.
P=W sinα + μN.
=W sinα +μW cosα.
=W (sinα+μcosα) {μ=mu i.e. coefficient of friction}.
P=W sinα + μN.
=W sinα +μW cosα.
=W (sinα+μcosα) {μ=mu i.e. coefficient of friction}.
(20)
Naresh Chaudhary said:
6 years ago
Hello, everyone, I have a question for everyone. Why the presence of friction converts mechanics into discontinuous nonlinear problems?
(2)
HARSHAVARDHAN said:
8 years ago
Required effort to move the block = P.
Resisting forces = (a component of the weight in the direction to opposite of P) + (friction force in opposite direction of moving).
= ( Wsina) + (u*Fn) (where Fn is the normal force acting on body)
= ( Wsina) + (u*Wcosa).
Thus for equilibrium,
P = Wsina + W*u*cosa.
P = W ( sina + ucosa).
Resisting forces = (a component of the weight in the direction to opposite of P) + (friction force in opposite direction of moving).
= ( Wsina) + (u*Fn) (where Fn is the normal force acting on body)
= ( Wsina) + (u*Wcosa).
Thus for equilibrium,
P = Wsina + W*u*cosa.
P = W ( sina + ucosa).
(2)
Ahsan said:
6 years ago
Thanks @Saddam Husain.
(1)
Rohit said:
9 years ago
From free body diagram
p = f + wsina {a=alpha}
= un + wsina
= uwcosa + wsina
= w(ucosa + sina)
So, the answer is c.
p = f + wsina {a=alpha}
= un + wsina
= uwcosa + wsina
= w(ucosa + sina)
So, the answer is c.
K. Ajaykumar said:
5 years ago
Thanks all for explaining.
Rambo said:
7 years ago
Can someone explain how sin, cos and tan is applied? Please tell me.
Fashir said:
7 years ago
Explanations are good. Thanks all.
Shubham singh kurvanshi said:
8 years ago
Here not mention the body will move upward or downward.
Whenever body is moving downward the effort will be (tan$cos@-sin@)W. And when the body moves upward the effort will be (tan$cos@+sin@)W.
So, option c is correct.
Whenever body is moving downward the effort will be (tan$cos@-sin@)W. And when the body moves upward the effort will be (tan$cos@+sin@)W.
So, option c is correct.
Ashif said:
8 years ago
Why not?
F=W tan (π+ α).
F=W tan (π+ α).
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