Chapter 2: Force systems Force : vector quantity (magnitude and - - PowerPoint PPT Presentation

Chapter 2: Force systems

Force: vector quantity (magnitude and direction are important) Unit: newton (N) Deformable body: point of action is also important! (fixed vector) Principle of Transmissibility

F r F r F r F r F r

Rigid body: force acting at any point on the line of action brings the same effect (sliding vector)

External and internal effects

Depend on body in consideration External Internal

F r F r

Fixed

F r

Concurrent forces

A

1

F r

2

F r R r A

2

F r

1

F r R r A

1

F r

2

F r R r

• Concurrent forces:

Forces that their line of action intersect at a point (point A)

• Their sum is called resultant R
• Line of action of resultant R must pass point A

Vector components and projections

a b

R r

1

F r

2

F r

a b

b

F r R r

a

F r

= orthogonal projection

• f (onto axis a and b)

b a F

F r r , R r

2 1, F

F r r R r = vector components

• f (along axis a and b)

2 1

F F R r r r + =

b a

F F R r r r + ≠

A special case of vector addition

2

F

1

F

1

R r

2

R r

2

F

1

F F F

R r

1

R r

2

R r

1

R r

2

R r

2

F

1

F F F

F1 and F2 are parallel Add F and –F to produce R1 and R2 Magnitude, direction and line of action of R can be known

r r r r r r r

Sample problem (1)

Combine the two forces P and T, which act on the fixed structure at B, into a single equivalent force R.

Sample problem (2)

Determine the components of the 800- N force F along the oblique axes a and

• b. Also, determine the projections of F
• nto the a- and b- axes

Sample problem (3)

The tension in cable AC is 8 kN. Determine the required tension T in cable AB such that the net effect of the two cable tension is a downward force at point A. Determine the magnitude R of this downward force.

Rectangular components

x y j ˆ i ˆ

θ

F r

x

F r

y

F r

y x

F F F r r r + = j F i F F

y x

ˆ ˆ + =

2 2 y x

F F F + = ) cos(θ F Fx = ) sin(θ F Fy =

1

tan ( / )

y x

F F θ

−

=

F F F

y x

r r r

• f

components vector , = F F F

y x

r

• f

components scalar , =

Reference axis

x y

F r

θ

Fx= Fcosθ Fy= −Fsinθ x y

F r

θ

Fx= −Fcosθ Fy= −Fsinθ y

F r

θ

x

β

Fx= Fsin(β−θ) Fy= −Fcos(β−θ)

Reference axis: assignment based on convenience

Vector summation

1

F r

2

F r

y x F2y Ry F1y F1x F2x Rx

• R

r

) ˆ ˆ ( ) ˆ ˆ (

2 2 1 1 2 1

j F i F j F i F F F R

y x y x

+ + + = + = r r r

j F F i F F j R i R

y y x x y x

ˆ ) ( ˆ ) ( ˆ ˆ

2 1 2 1

+ + + = +

∑

= + =

x x x x

F F F R

2 1

∑

= + =

y y y y

F F F R

2 1

Sample problem (4)

If the equal tensions T in the pulley cable are 400 N, express in vector notation the force R exerted on the pulley by the two

• tensions. Determine the magnitude of R.

Sample problem (5)

If the cable tension is 750 N, determine the n- and t-components of this force acting on point A of the bar.

Sample problem (6)

(1) Write F in terms of the unit vector i and j and identify both its vector and scalar components. (2) Determine the scalar components of the force vector F along the x'- and y'- axes. (3) Determine the scalar components of F along the x- and y'- axes. The 500-N force F is applied to the vertical pole as shown.