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.
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)
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.
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.
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).
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)
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.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).
Shankar said:
9 years ago
What is the criteria to decide force due to gravity parallel to inclined plane wsin (a)?
Raj said:
9 years ago
Force due to gravity parallel to inclined is w(sina) force due to friction is u(w cosa).
Post your comments here:
Quick links
Quantitative Aptitude
Verbal (English)
Reasoning
Programming
Interview
Placement Papers