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.
Sagar said:
1 decade ago
Consider the free body of diagram of the block moving upwards. The force P we are applying to the right and component of weight along the inclined plane wsina to the left and frictional force uwcosa to the left. Consider the force balance we will arrive at the equation as stated at C.
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).
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).
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).
Yogesh patel said:
9 years ago
Thanks for guiding me.
Ankit ughade said:
9 years ago
Great @Diwakar.
Raghu said:
9 years ago
Thanks for your guidance.
Manishankar said:
9 years ago
Anyone provide me, its free body diagram.
Tanu said:
9 years ago
What is free body diagram? Please explain.
Raj said:
9 years ago
Force due to gravity parallel to inclined is w(sina) force due to friction is u(w cosa).
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