Moment is the turning or rotating effect of a force about a point. It is calculated by multiplying the force by the perpendicular distance from the point to the line of action of the force. A couple is formed by two equal, unlike parallel forces with zero net force but a constant moment. Varignon's theorem states that the moment of the resultant force is equal to the sum of the moments of the individual forces about the same point. Systems of forces and couples can be replaced by an equivalent single force system using this theorem.
Hooke's law sample problems with solutionsBasic Physics
A graph of force (F) versus elongation (x) shown in figure below. Find the spring constant!
Solution
Hooke's law formula :
k = F / x
F = force (Newton)
k = spring constant (Newton/meter)
x = the change in length (meter)
Hooke's law sample problems with solutionsBasic Physics
A graph of force (F) versus elongation (x) shown in figure below. Find the spring constant!
Solution
Hooke's law formula :
k = F / x
F = force (Newton)
k = spring constant (Newton/meter)
x = the change in length (meter)
This presentation explains about the introduction of Bode Plot, advantages of bode plot and also steps to draw Bode plot (Magnitude plot and phase plot). It explains basic or key factors used for drawing Bode plot. It also explains how to determine Magnitude, phase and slope for basic factors. It also explains how to determine stability by using Bode Plot and also how to determine Gain Crossover Frequency and Phase Crossover Frequency, Gain Margin and Phase Margin. It also explains drawing Bode plot with an example and also determines stability by using Bode Plot and also determines Gain Crossover Frequency and Phase Crossover Frequency, Gain Margin and Phase Margin.
In this presentation we can get to know how we can construct a bode plot with suitable examples Of different -different orders.
Along with that a simulation model on MATLAB with graph.
And in this we have explained about the transfer function, poles & zeroes.And the basic concept of stability.
This presentation explains about the introduction of Bode Plot, advantages of bode plot and also steps to draw Bode plot (Magnitude plot and phase plot). It explains basic or key factors used for drawing Bode plot. It also explains how to determine Magnitude, phase and slope for basic factors. It also explains how to determine stability by using Bode Plot and also how to determine Gain Crossover Frequency and Phase Crossover Frequency, Gain Margin and Phase Margin. It also explains drawing Bode plot with an example and also determines stability by using Bode Plot and also determines Gain Crossover Frequency and Phase Crossover Frequency, Gain Margin and Phase Margin.
In this presentation we can get to know how we can construct a bode plot with suitable examples Of different -different orders.
Along with that a simulation model on MATLAB with graph.
And in this we have explained about the transfer function, poles & zeroes.And the basic concept of stability.
Unit 1. force system, solved problems on force system.pdfVrushali Nalawade
Solved problems on the Force system
engineering mechanics
applied mechanics
force
numericals for practice
parallelogram law
law of moment
moment
couple
varignon's theorem
triangle law
resultant force
magnitude
direction
composition and resolution
perpendicular component
non-perpendicular component
moment of force
force system
method of resolution
A Simplified Approach to Calculating Truss ForcesJames Nadir
A simplified approach to calculating truss forces. This presentation is built upon a wonderful presentation which comes from Fazirah Abdul Ghafar. This presentation uses Power Point Animations and is best viewed using Power Point Slide Show to present.
1. RESULTANT OF NON-CONCURRENT FORCE SYSTEM
Moment and Couple
Moment of a force means turning or rotating effect of the force on the body. Magnitude of moment is
obtained by multiplying perpendicular distance (from the particular point on the line of action of the force)
and the force.
Moment of a force F about point A can be denoted as MA. Hence, MA = d1 . F (clockwise) while MB = 0 and
MC = d2 . F (counter clockwise). If the point under consideration lies on the line of action force, its moment
will be zero or the body will only tend to translate without rotating effect.
Two equal, unlike parallel forces form a couple. It has two important characteristics. (1) Resultant (or sum
of components of forces) force is zero and (2) moment of a couple always remains constant. Therefore at
what point a couple applied has absolutely no consequence.
Difference between Moment and Couple
‘Moment of a force’ may have different magnitude and different direction about different points in the
same plane, while ‘couple’ (moment of couple) has same effect anywhere on the body irrespective of the
point on it; or its moment is constant.
Moment of a force cannot have ‘zero resultant’ but couple has resultant force zero. Moment of a couple
can never be zero while if the point is on the line of action of the force, moment of force can be zero.
Varignon’s Theorem of Moments
Statement: The moment of resultant of a force system is equal to algebraic sum of moments of all
components or forces in the system about the same point. Hence, mathematically,
d ⋅ R = (d1 ⋅ F1 ) + (d2 ⋅ F2 ) + ...
2. Example:
Replace the given system of forces acting on the beam AB shown in the figure, by (A) resultant and
distance at which it acts from point A, (B) an equivalent force-couple system at A, and (C) an equivalent
force-couple system at B.
Solution:
(A) Resultant and distance from point A
The resultant is: ΣR = (300 − 1200 + 200 − 500) N
R = 1200 N ↓
The distance from point A, using Varignon’s Theorem of Moments (VTM), CW positive:
(1200 N) x = −(300N)(2m) + (1200N)(5m) − (200N)(7m) + (500N)(11m)
x = 7.92 m
Equivalent force system:
(B) Single force couple system from point A
For replacing the given system by force couple system at A, find the net (resulting) moment of forces
at point A. Assume forces in CW rotation with respect to point A as positive.
ΣMA = −(300N)(2m) + (1200N)(5m) − (200N)(7m) + (500N)(11m)
ΣMA = 9500 N – m clockwise
3. (C) Single force couple system from point B
Similarly at point B, assume forces in CW rotation with respect to point B as positive.
ΣMB = −(500N)(1m) + (200N)(5m) − (1200N)(7m) + (300N)(10m)
ΣMA = -4900 N – m counter clockwise
Example:
Replace the system of forces and couple by a single force system at A.
Solution:
4
( )
ΣR X = −75 N − 100 N = −155 N ←
5
3
ΣR Y = −200 N + 50 N − (100 N) = −210 N
↓
5
Resultant Force:
R = (− 155 N)
2
(
+ − 210 N )2
R = 261 N
Direction:
− 210 N
θ = tan −1
− 155 N
θ = 53.57o
4. Equivalent moment about A, CW positive:
3
ΣM A = −(50N)(2m) − (80Nm) + (100N) (4m)
5
ΣMA = 60 N – m clockwise
Equivalent force system:
Example:
If resultant moment at A of three coplanar forces is 88.4 N-m Clockwise, determine F. What is the
resulting force of the system?
Solution:
4
ΣMA = 88.4 N − m = (100N sin 30o )(0.8m) + (F) (1.6m) − (90N)(0.6m)
5
F = 80 N
3
ΣR X = −100N cos 30o − 90N + (80N) = −38.603 N
5 ←
4
ΣR y = −100N sin 30o − (80N) = −114 N
5 ↓
Resultant Force:
R = 120.359 N
Direction: θ = 71.29o