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FORCES

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TYPES OF FORCES

Forces: A variety of forces are investigated to show that forces can change

either the shape of an object or its motion. Both balanced and unbalanced

forces are considered.

Types: Internal and External forces

Examples: Tension (stretch) AND Compression (crush)

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Bending Tension Torsion (twisting)

Constant Force

Gravitational pull

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External/Internal forces

External force = stresses that act on the structure from the outside ofthe structure. External force produce internal forces. Example: gravity

Internal force = reaction forces that act from the inside of the material.

Example: compression, tension and shear forces.

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Classification of forces

•

Colinear: Forces acting along the same line of action. Themagnitude of a single equivalent force is the same as the sum

of the colinear forces

•

Concurrent: Pass through the same point in space

• Coplanar: Lie in the same plane

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Coplanar force = all the forces acting in in one plane.

They may be concurrent, parallel, non-concurrent or non-parallel.

All of these systems can be resolved by using graphic statics or algebra.

A parallel coplanar force system consists of two or more

forces whose lines of action are ALL parallel.

These vectors do not meet at a single point (non-

concurrent) and none of them are parallel (non-parallel).

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Resultant

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COMPOSITION AND RESOLUTION OF FORCES

It is possible to find a single force, which will have the sameeffect as that of a number of forces acting on a body

Such a single force is called Resultant force and the process or

method of finding out the resultant force is called composition of forces

If several coplanar (lie in the same plane) forces F1, F2, F3, . . .,

are applied to a body at one point, they represent a system of

forces that can be reduced to a single resultant force. It then

becomes possible to find this resultant by successive applications

of the parallelogram law.

REVISE = vectors!!!

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Successive applications of parallelogram law:

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For example, the four forces F1, F2, F3, and F4 acting on a body at point

A (Fig. 1a).

To find their resultant (R), we begin by obtaining the resultant AC of

the two forces F1 and F2. Combining this resultant with the force F3, weobtain the resultant AD, which must be equivalent to F1, F2, and F3.

Finally combining the forces AD and F4, we obtain the resultant R of the

given system F1-F4.

The same resultant R will be obtained by successive geometricaddition as shown above in Fig.1b.

In this case we begin with the vector AB representing the force F1.

From the end B of this vector we construct the vector BC, representing

the force F2

, and afterward, the vectors CD and DE representing theforces F3 and F4 respectively.

In the polygon ABCDE obtained in this way, the vector AE gives the

resultant R.

The polygon ABCDE is called the polygon of forces.

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Free Body Diagram (FBD)

A graphical sketch of the system showing all external/reaction forces and moments applied to the

system.

Benefits:

1)Provides a catalog of all force/moments acting on

the particle

2) Provides a graphical display of known information

(force/moment magnitudes, directions and line of

actions)

3)Provides a record of the geometric dimensions

needed for establishing moments of the forces

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If we have a set of concurrent forces (pass

through same point), we can resolve these

forces into a single force

To compute the resultant of several concurrent

forces:

• Determine the angle of each force with

respect to x and/or y axis• Find x and y components for each force

• Find sum of colinear forces (Fx, Fy, Fz)

REVISE = vectors!!!

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Composition of Forces by Method of Resolution

Let P1, P2, P3, and P4 are the forces of a system as shown. Components of

these forces are shown in x- and y- direction

∑X = P 1x + P 2x + P 3x + P 4x

∑ Y = P 1y + P 2y + P 3y + P 4y

Magnitude, R = [( ∑ X)2 + ( ∑ Y)2]1/2

(rearranged from R² = ΣX² + ΣY²)

And its inclination to x-axis is given by:

θ = tan-1 ( ∑ Y /∑ X )

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Example 1 – Concurrent, Coplanar ForcesIn this simple structure, a 500 lb. load is supported by two cables, which in turn

are attached to walls. Determine the forces (tensions) in each cable.

From Diagram-1, it is clear that, this problem may be classified as a problem

involving Concurrent, Coplanar Forces.

Concurrent: The vectors representing the two support forces in Cable 1 and

Cable 2, and the vector representing the load force all intersect at one point

(Point C, See Diagram 2).

Coplanar: This is a 2-D problem, that is forces lie in the x-y plane only.

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1. Draw a Free Body Diagram (FBD) of the structure or a portion of the structure. This

Free Body Diagram should include a coordinate system and vectors representing all the

external forces (which include support forces and load forces) acting on the structure.

These forces should be labeled either with actual known values or symbols

representing unknown forces.2. Diagram 2 is the Free Body Diagram of point C with all forces acting on point C shown

and labeled.

3. Resolve (break) forces into their x and y components. Notice that T1, and T2, the

vectors representing the tensions in the cables are acting at angles with respect to the x-

axis, that is, they are not simply in the x or y direction. Thus for the forces T1, T2, we must

replace them with their horizontal and vertical components. In Diagram 3, the

components of T1 and T2 are shown.

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Since the components of T1 and T2 (T1 sin 53o, T1 cos 53o, T2 sin 30o, T2 cos 30o) are

equivalent to T1 and T2, in the final Diagram 4, remove T1 and T2 which are nowrepresented by their components. Notice that we do not have to do this for the load force

of 500 lb., since it is already acting in the y-direction only.

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4. Apply the Equilibrium Conditions and solve for unknowns. In this step, apply the

actual equilibrium equations. Since the problem is in 2-D only (coplanar) we have the

following two equilibrium conditions:

The sum of the forces in the x direction, and the sum of the forces in the y directionmust be zero.

5. We now place our forces into these equations, remembering to put the correct +ve

and –ve sign with the force.

= ????

= ????

ANS: T1 = 436 lb.; T2 = 303 lb

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Draw the FBD of an object sliding along a

linear x-axis with a force of 100 N.

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EXAMPLE: A block of mass “m” is at rest on a rough incline of angle “θ”. Find

the contact forces and the net contact force on the block.

A block on an incline Forces on the block

The forces on the block are: (i) its weight; (ii) normal force; and (iii) Friction force. These

forces are not concurrent (see above figure).

However, no turning effect is involved. We can, therefore, treat force system concurrent

with the "center of mass" of the block. In order to analyze the forces, we consider a

coordinate system as shown in the figure.

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Free body diagram

Forces are shown with block as a

point with concurrent forces

Since the block is at rest, the forces on the block are balanced :

∑Fx=mgsinθ−Ff =0∑Fy=N−mgcosθ=0

There are two contact forces (i) normal force and (ii) friction. The friction is given by the first

equation : Ff =mgsinθ

The normal force is given by the second equation : N=mgcosθ

The net contact force is vector sum of two contact forces,Fc = N + Fr

Hence, magnitude of net contact force is :

FC = [ (N2+Ff 2)]1/2

= [(mg cosθ)² + (mg sinθ)²]

=[ (mg) 2(cos2θ+sin2θ) ]1/2

FC =mg

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