Resolution of forces
Mechanics
It is defined as that branch of science, which describes and predicts the conditions of rest or motion of bodies under the action of forces. Engineering mechanics applies the principle of mechanics to design, taking into account the effects of forces. Statics deal with the condition of equilibrium of bodies acted upon by forces.
It is defined as that branch of science, which describes and predicts the conditions of rest or motion of bodies under the action of forces. Engineering mechanics applies the principle of mechanics to design, taking into account the effects of forces. Statics deal with the condition of equilibrium of bodies acted upon by forces.
Rigid bodyA rigid body is defined as a definite quantity of matter, the parts of which are fixed in position relative to each other. Physical bodies are never absolutely but deform slightly under the action of loads. If the deformation is negligible as compared to its size, the body is termed as rigid.
Force
Force may be defined as any action that tends to change the state of rest or motion of a body to which it is applied.
The three quantities required to completely define force are called its specification or characteristics. So the characteristics of a force are:
1. Magnitude
2. Point of application
3. Direction of application
The line of action of force
The direction of a force in the direction, along with a straight line through its point of application in which the force tends to move a body when it is applied. This line is called the line of action of force.Representation of force
Graphically a force may be represented by the segment of a straight line.
Various Systems of Forces
Parallel forces on a plane
1. Like parallel forces: Coplanar parallel forces when acting in the same direction.
2. Unlike parallel forces: Coplanar parallel forces when acting in opposite direction.
3. Resultant of like parallel forces: Let P and Q are two like parallel forces act at points A and B.
R = P + Q 
4. Resultant of unlike parallel forces:
R = P - Q 
Couple Force
Two, unlike equal parallel forces, form a couple.
The rotational effect of a couple is measured by its moment. Moment = P × l
Sign convention: Anticlockwise couple (Positive)
Clockwise couple (Negative)
The rotational effect of a couple is measured by its moment. Moment = P × l
Sign convention: Anticlockwise couple (Positive)
Clockwise couple (Negative)
Question 1: A rigid bar CABD supported as shown in the figure is acted upon by two equal horizontal forces P applied at C and D. Calculate the reactions that will be induced at the points of support. Assume l = 1.2 m, a = 0.9 m, b =0.6 m.
Solution: Taking moment about A,
Ra = Rb
Rb ✖l + P ✖ b = P ✖ a
Rb = P(0.9 - 0.6) ➗ 1.2
Rb = 0.25P(↑)
Ra = 0.25P(↓)
Question2: Owing to weight W of the locomotive shown in the figure, the reactions at the two points of support A and B will each be equal to W/2. When the locomotive is pulling the train and the drawbar pull P is just equal to the total friction at the points of contact A and B, determine the magnitudes of the vertical reactions Ra and Rb.
Solution
åV = 0Ra + Rb = W
Taking moment about B,
å MB = 0
Ra ´ 2a + P ´ b = W ´ a
Þ Ra =(W .a - P.b)
2a
Solution
åV = 0
Ra + Rb = 3P + Q
Þ Ra + Rb = 3´ 90 + 72
Þ Ra + Rb = 342KN
å M A = 0
Rb ´ 9.6 = 90 ´1.8 + 90 ´ 3.6 + 90 ´ 5.4 + 72 ´ 8.4
Þ Rb = 164.25KN
\ Ra = 177.75KN
Question 4: The beam AB in figure is hinged at A and supported at B by a vertical cord which passes over a friction-less pulley at C and carries at its end a load P. Determine the distance x from A at which a load Q must be placed on the beam if it is to remain in equilibrium in a horizontal position. Neglect the weight of the beam.
Solution
Take å M A = 0
S ´ l = Q ´ x
Þ x = P.l Q
Composition of force
The reduction of a given system of forces to the simplest system that will be its equivalent is called the problem of the composition of forces into components.
These components are also referred to as rectangular components of a force.
To find the component of a vector along a given axis, we drop a perpendicular on the given axis from the vector.
For example, OA is the given vector. We have to find its component along the horizontal axis. Let us call it the x-axis. We drop a perpendicular AB from An onto the x-axis. The length OB is the component of OA along the x-axis. If OA makes angle p with the horizontal axis, then in triangle OAB, OB/OA = Cos P or OB = OA Cos P.
Remember that component of a vector is a scalar quantity. If the component is along the negative direction, we put a (-) sign with it.)
Usually, we resolve the vector into components along mutually perpendicular components.
OB is the x component OB = OA Cos p.
Similarly component along the vertical direction or the y-axis is OC
OCAB is a rectangle.
So OC = AB
look at triangle OAB again,
AB/OA = Sin p
=> AB = OA Sin p = OC
Thus y component OC = OA Sin p.
Note that p is the angle with the horizontal axis.
Solved Numericals on "Rectangular components of a Force"
Question
Find the X and Y components of a 25 m displacement at an angle of 210 deg.
Answer
OA is the displacement vector.
The angle with the horizontal axis is 210 deg - 180 deg = 30 deg
x component = OB = -25 Cos 30 deg = -21.7
y component = AB = -25 Sin 30 deg = -12.5 m
Note that each component is pointing along the negative coordinate direction and thus we must take it as negative.
Now we will solve a problem using the component method
Question
Find the resultant of the following two displacements: 2 m at 30 deg and 4 m at 120 deg. The angles are taken relative to the x-axis.
Answer
Rx = 2 Cos 30 deg - 4 Cos 60 deg = - 0.268 m
Ry = 2 Sin 30 deg + 4 Sin 60 degg = 4.46 m
R = √(Rx2 + Ry2)
= √(-0.2682 + 4.462) = 4.47 m
tan q = Ry/Rx = 4.46/0.268
=> q = 86.6 deg
p = 180 deg - 86.6 deg = 93.4 deg
Parallelogram law
If two forces represented by vectors OA and OB acting under an angle α are applied to a body at point O. Their action is equivalent to the action of one force, represented by vector OC, obtained as the diagonal of the parallelogram constructed on the vectors OA and OB directed as shown in the figure.
Force OC is called the resultant of OA and OB and the forces are called its components.
Force OC is called the resultant of OA and OB and the forces are called its components.
Special cases
Question
Two forces of 3 N and 4 N are acting at a point such that the angle between them is 60 degrees. Find the resultant force
Answer
Magnitude R of the resultant force is R = √(32 + 42 + 2 x 3 x 4 Cos 60 deg)
= √(9 + 16 + 12) = √(37 = 6.08 N
Direction of R is given by finding the angle q
tan q = (3 Sin 60 deg)/(4 + 3 Cos 60 deg) = 0.472
q = tan-1 0.472
= 25.3 deg
Thus R is 6.08 N in magnitude and is at an angle of 25.3 deg to the 4 N force.
Question
A car goes 5 km east 3 km south, 2 km west and 1 km north. Find the resultant displacement.
Answer
First, we will make the vector diagram
O to A 5 km east
A to B 3 km south
B to C 2 km west
c to D 1 km north
Net displacement is OD
Along the horizontal direction: 5 km east - 2 km west = 3 km east
Along the vertical direction: 3 km south - 1 km north = 2 km south
OD = √(32 + 22 + 2 x 3 x Cos 90 deg)
= √(32 + 22)
= 3.6 km
tan p = 2/3
or p = tan-12/3 = 34 deg
Thus resultant displacement is 3.6 km, 34 deg south of east.
Action and reaction
Often bodies in equilibrium are constrained to investigate the conditions.
Free body diagram
Free body diagram is necessary to investigate the condition of equilibrium of a body or system. While drawing the free body diagram all the supports of the body are removed and replaced with the reaction forces acting on it.
Draw the free body diagrams of the following figures.
Draw the free body diagrams of the following figures.
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3. Draw the free body diagram of the following figures. |
Equilibrium of colinear forces:
Two forces can be in equilibrium only if they are equal in magnitude, opposite in direction and collinear in action.
Figure: Tension
Figure: Compression
Equilibrium of concurrent forces in a plane
- If a body known to be in equilibrium is acted upon by several concurrent, coplanar forces, then these forces or rather their free vectors, when geometrically added must form a closed polygon.
- This system represents the condition of equilibrium for any system of concurrent forces in a plane.
Ra w tan S w sec
Lami’s theorem
If three concurrent forces are acting on a body kept in an equilibrium, then each force is proportional to the sine of the angle between the other two forces and the constant of proportionality is same.
Problem: An electric light fixture of weight Q = 178 N is supported as shown in the figure. Determine the tensile forces S1 and S2 in the wires BA and BC, if their angles of inclination are given.
Solution:
Find S1 and S2 to get Answer.
Theory of transmissibility of a force
The point of application of a force may be transmitted along its line of action without changing the effect of force on any rigid body to which it may be applied.
Moment of a Force
- Considering wrench subjected to two forces P and Q of equal magnitude. It is evident that force P will be more effective compared to Q, though they are of equal magnitude.
- The effectiveness of the force as regards it is the tendency to produce rotation of a body about a fixed point is called the moment of the force with respect to that point.
- Moment = Magnitude of the force × Perpendicular distance of the line of action of force.
- Point O is called the moment centre and the perpendicular distance (i.e. OD) is called moment arm.
- Unit is N.m
Theorem of Varignon
The moment of the resultant of two concurrent forces with respect to a centre in their plane is equal to the algebraic sum of the moments of the components with respect to some centre.
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