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 2 of 3.
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.
Raj said:
9 years ago
Force due to gravity parallel to inclined is w(sina) force due to friction is u(w cosa).
Tanu said:
9 years ago
What is free body diagram? Please explain.
Manishankar said:
9 years ago
Anyone provide me, its free body diagram.
Raghu said:
9 years ago
Thanks for your guidance.
Ankit ughade said:
9 years ago
Great @Diwakar.
Yogesh patel said:
9 years ago
Thanks for guiding me.
K.Tamil selvan said:
10 years ago
From free body diagram:
ΓH = 0;
p = wcosa+MR.
Γv = 0;
R = wsina.
p = wcosa+Mwsina.
p = w(cosa+Msina).
ΓH = 0;
p = wcosa+MR.
Γv = 0;
R = wsina.
p = wcosa+Mwsina.
p = w(cosa+Msina).
Diwakar Pandey said:
1 decade ago
Effort applied = P (acting to the right).
Frictional force = uWcosa(acting against P).
Gravitational force = Wsina(acting downwards).
So equation will be in equilibrium as:
P = Wsina + uWcosa.
P = W(sina+ ucosa).
Frictional force = uWcosa(acting against P).
Gravitational force = Wsina(acting downwards).
So equation will be in equilibrium as:
P = Wsina + uWcosa.
P = W(sina+ ucosa).
Saddam husain said:
1 decade ago
From free body diagram,
p = u*R+w*sina.
p = u*(w*cosa)+wsina.
p = w(ucosa+sina).
p = u*R+w*sina.
p = u*(w*cosa)+wsina.
p = w(ucosa+sina).
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