5.2 Rigid Bodies in Two-Dimensional Force Systems

5.2 Rigid Bodies in Two-Dimensional Force Systems Example 1, page 1 of 1

The force components at A can be assumed to act in any

direction we choose. Choosing them in the direction of the

inclined coordinate system simplifies the calculation a little.

40 lb

B y

A x

2

A y

y

3 ft

3 ft

40 lb

Free-body diagram of rod AB.

1. Determine the reactions at A and

B. The weight of the rod is

negligible.

1

3 ft

x

3

A

4

Solving these equations simultaneously gives

A x = 0 Ans.

Ay = 20.0 lb Ans.

By = 20.0 lb Ans.

+5

+

+

Equilibrium equations for rod AB:

F x = 0: A

x = 0

Fy = 0: A y 40 lb + B

y = 0

MA = 0: 40 lb)(3 ft) + B y(3 ft + 3 ft) = 0

The force from the roller must be perpendicular

to the plane upon which the roller rests.

3 ft B

30°

30°

A

B


A

B

3 kip

Free-body diagram of member AB1

4 ft4 ft

2. Determine the reactions at A and B. The

weight of the rod is negligible.

4 ft 4 ft

3 kip

B y

5 ft

B x

The force from the roller at A

must be perpendicular to the plane

upon which the roller rests.

2

Equilibrium equations for member AB:

Fx = 0: A x + B

x = 0

Fy = 0: 3 kip + B y = 0

MB = 0: A x(5 ft) + (3 kip)(4 ft) = 0

4

+

+

+

The pin support at B

provides two force components.

3


A x = 2.4 kip Ans.

B x = 2.4 kip Ans.

B y = 3.0 kip Ans.

5

A

Ax

B

5 ft


3. Determine the reactions at supports A and C.

1 Free-body diagram of member ABC

2 The force from the roller

is perpendicular to the

inclined plane. The

angle will be

determined later.

3 The pin support provides two

components of reaction.B

400 lb

A

C

F A

C y

C x

30°

4 ft

5 ft

B

400 lb

A

C


Equilibrium equations for member ABC:

Free-body diagram showing horizontal and vertical components

of the unknown reaction force F A.

F A sin

5

4

F A cos

A

C y

C

C x

F x = 0: F

A cos + C x = 0 (1)

F y = 0: F

A sin + C y 400 lb = 0 (2)

MC = 0: F A cos (4 ft) + (400 lb)(5 ft)

F A sin 5 ft 4 ft)/tan 30°] = 0 (3)

+

+

+

30°

400 lb

B

5 ft(4 ft)/tan 30°

4 ft


C y = 260 lb

C

|C x| = 81.1 lbB

400 lb

60°

A

F A = 162.2 lb

60°

30°

30°

Substituting = 60° in Equations 1, 2, and 3

and solving gives

F A = 162.2 lb Ans.

C x = 81.1 lb Ans.

C y = 260 lb Ans.

We had assumed that C x pointed to the right, but

solving the equilibrium equations gave a negative

value for C x, so our assumption was wrong. C

x

points to left as shown.

Free-body diagram showing forces

with correct senses

9

8

Geometry

7

6

B

A

= 90° 30 ° =


4. The automobile shown is being pushed up the incline at

a constant velocity by a force, P, from a tow truck. The

2,600-lb weight of the car acts at the center of gravity, O,

and the friction forces acting on the wheels are negligible.

Determine the force P and the forces from the ground

acting on the individual tires.

1 Free-body diagram of car

2 It is convenient to use x and y coordinates that

are parallel and perpendicular to the incline,

because the forces from the tow truck and

from the ground acting on the tires are parallel

and perpendicular to the incline.

3 If F B is the force acting on each front

wheel, then 2F B is the resultant force

acting on the body.

55 in.

O

12 in.

20 in.

A

B

O

A

B

y

x

2 F A

2 F B

2,600 lb

P

P

15 in.

10°


Solving simultaneously, with = 80°, gives

P = 451 lb Ans.

F A = 300 lb Ans.

F B = 980 lb Ans.

= 90° 10° = 80°

Geometry5

P (2,600 lb) cos = 0

2F A + 2F

B (2,600 lb) sin = 0

P(8 in.) 2F A(55 in.) + 2F

B(15 in.) = 0

Equilibrium equations for the car:

F x = 0:

F y = 0:

M O = 0:

+

+

+

6

10°

20 in 12 in = 8 in

15 in.

P

55 in.

2,600 lb

2 F B

2 F A

y

B

A

O

Free-body diagram of car, with dimensions shown4

x

10°20 in.

12 in.


5. Plate ABCD is supported by cord EC and a pin at A.

Determine the tension in the cord and the horizontal and

vertical reactions at A. The weight of the plate is negligible.

4 kN D

AB

C

E

5 kN·m

20°

3 m

2 m


B

4 kN D

C

A y

A x

T

A5 kN·m

4 The 5 kN·m couple-moment at corner B can be

considered to act anywhere; in particular, then, we can

consider it to act at point A, about which we are

summing moments. So we just insert "5 kN·m" into the

equation. The minus sign signifies that the 5 kN·m

couple-moment has a sense opposite to what we have

chosen to call a positive sense.

20°

3 m

2 m

1 Free-body diagram of plate ABCD

2 The tension T points away from the

plate and is collinear with the cord.

3 Equilibrium equations for plate ABCD

F x = 0:

Fy = 0:

MA = 0:

+

+

+

A x + 4 kN T cos 20° = 0

A y + T sin 20° = 0

4 kN)(2 m) + (T cos 20°)(2 m) + (T sin 20°)(3 m) 5 kN·m = 0


Initially we arbitrarily assumed A y points upward.

Solving the equilibrium equations gave A y = 1.53 kN.

The minus sign shows that our assumption was wrong,

and the correct sense of Ay is down, as shown on this

free body.

7

C

A

A y = 1.53 kN

A x = 0.20 kN

B

5 kN · m

Free-body diagram of plate ABCD showing

correct senses of forces.

4 kN D

6

T = 4.47 kN


T = 4.47 kN Ans.

A x = 0.20 kN Ans.

A y = 1.53 kN Ans.

5

20°


O

A

B

C

60 lb

3 ft

45°

6. Member ABC has the shape of a quarter circle centered at O.

Determine the reactions at each of the roller supports.

45°


Free-body diagram of member ABC

The force from the

roller must be

perpendicular to the

plane upon which

the roller rests. This

is true at all three

points, A, B, and C.

F x = 0:

F y = 0:

M O = 0:

60 lb F B sin 45° = 0

F A + F

B cos 45° F C = 0

(60 lb)(3 ft) F C(3 ft) = 0

Equilibrium equations for plate member ABC

1

2 3


FA = 0 Ans.

FB = 84.9 lb Ans.

FC = 60.0 lb Ans.

4

60 lb

O

F A

F B

F C

A

B

C

45°

3 ft

The lines of action of the forces FA

and FB go through point O, so FA

and FB have no moment about O.

+

+

+


6 kip

C

B

AE F

D

2 ft

5 ft

2 ft

8 ft 12 ft

7. The end of pole DA is buried deep in the ground. The

tension in cable BE is 2 kip and in cable CF is 3 kip.

Determine the reactions at A.


D6 kip

C

B T CF = 3 kip

T BE = 2 kip

MA

x

A y

1

2

3

4

5

C

B

A

E

F

( ) = tan-1

= tan-1 ( )

= 58.00°

= 59.74°

2 ft

5 ft

2 ft

2 ft

5 ft

8 ft

Free-body diagram of pole ABCD

The tensions

point away from

the pole and are

collinear with the

cables.

Because the end of the pole is buried

in the ground, not only do two force

components A x and A

y act at the end

of the pole, but also a couple moment

M acts to prevent the pole from

rotating.

Equilibrium equations for pole ABCD

Geometry

F x = 0:

F y = 0:

M A = 0:

+

+

+

A x (2 kip) sin + (3 kip) sin + 6 kip = 0

A y (2 kip) cos (3 kip) cos = 0

M + (2 kip)(sin )(5 ft) (3 kip)(sin (5 ft + 2 ft)

(6 kip)(5 ft + 2 ft + 2 ft) = 0

12 ft

A 8 ft5 ft

5 ft + 2 ft12 ft


Initially we assumed that A x acted to the

right. Solving the equilibrium equations

gave a negative value for A x, so the

correct sense of A x must be to the left as

shown here.

Free-body diagram showing the

correct sense of the forces.

Solving these equations, with = 58.00° and = 59.74°, gives

A x = 6.9 kip Ans.

A y = 2.6 kip Ans.

M = 63.7 kip·ft Ans.

6

6 kip

3 kip

2 kip

B

C

D

A y = 2.6 kip

A x = 6.9 kip

M = 63.7 kip·ft

7

8


8. A smooth slot has been cut in the plate shown, and

a pin, C, fixed to a support behind the plate, fits in

the slot. Determine the forces acting on the plate at

the supports A and B and from pin C.

2 lb

B

C

4 lb·ft

D

A

15°

3 ft 2 ft

1.5 ft

1.5 ft


The 4 lb·ft couple moment at corner D can be

considered to act anywhere; in particular, then, we

can consider it to act at point A, about which we are

summing moments. So we just insert "4 lb·ft" into

the equation.

5

4 lb·ft

A

C

B

D

2 lb

F BN

C

F A

Free-body diagram of plate ABD1

Because the slot is smooth, no

friction force acts on it from the pin.

The only force from the pin must be

the normal force, NC. That is, the

force is perpendicular to the slot.

Here its direction has been arbitrarily

assumed to be up and to the right.

2

The force from the roller supports at B and A

are perpendicular to the surface upon which

the roller rests.

3

2 lb F B + N

C cos = 0

F A + N

C sin = 0

4 lb·ft (N C cos )(1.5 ft ) + (N

C sin (3 ft)

+ F B (1.5 ft + 1.5 ft) = 0

Equilibrium equations for plate ABD:

F x = 0:

F y = 0:

M A = 0:+

+

4

+

3 ft 2 ft

1.5 ft

1.5 ft


= 90° 15° = 75°

A

F A = 2.94 lb

2 lb

F B = 1.212 lbB

N C = 3.04 lb

C

4 lb·ft

D

Initially we arbitrarily assumed N C points upward

and to the right. Solving the equilibrium

equations gave N C = 3.04 lb. The minus sign

shows that our assumption was wrong, and the

correct direction of N C is down and to the left as

shown on this free body.

9

C 15°

Free-body diagram of plate ABD showing correct senses of forces

Substituting the value = 75° into the equilibrium

equations and solving simultaneously, gives

F A = 2.94 lb Ans.

F B = 1.212 lb Ans.

N C = 3.04 lb Ans.

Geometry 7

8

6

15°


100 mm

AE

CD

B

12 N·m

9. Pin C is rigidly attached to the plate and can slide freely

in the slot cut in member DB. Determine the reaction

force at the pin support at A and the force transmitted at C.

425 mm

30°


Free-body diagram of plate ACE

Because the slot is smooth, no friction force acts on

it from the pin. The only force from the pin must

be the normal force, that is, the force is

perpendicular to the slot. Here its direction has

been arbitrarily assumed to be up and to the left.

Equilibrium equations for plate ACE

100 mm

E

12 N·m

C

N C

A x

A y

1

2

3

F x = 0:

F y = 0:

M A = 0:

+

+

+

A x N

C sin = 0

A y N

C cos = 0

12 N·m (N C sin (0.425 m) + (N

C cos )(0.1 m) = 0

D30°

425 mm


6 Free-body diagram of plate ACE showing correct

senses of forces

NC = 40.1 N

C

A x = 20.1 N

A y = 34.7 N

D

E

12 N·m

The sense of the force

vector has been reversed

because solving the

equilibrium equations

gives a minus sign.

7

30°

30°

A

Substituting = 30° into the equilibrium equations and

solving simultaneously gives

A x = 20.1 N Ans.

A y = 34.7 N Ans.

NC = 40.1 N Ans.

Geometry 54

C

D = 90° 60° = 30°

90° 30° = 60°


2 m 3 m2 m

3 kN·mA B D

C

10. Member ABD is supported by a pin at A and a cord attached at

B and D. The cord passes over a frictionless pulley at C.

Determine the reaction at A and the tension in the cord.

Moment equilibrium for pulley C: The moment about the

center produced by T 1 must equal the moment produced by

T 2. That is, T

1r = T 2r. Canceling the r's gives T

1 = T 2. In

fact, it's always true that for a frictionless pulley, the

tensions on the two sides of the pulley must be equal.

1

C

Radius r

T 1

T 2

5 kN

40°60°


Free-body diagram of beam ABD Because the tensions on the two sides

of pulley C are equal, the same tension

T acts on point B as on point D.

2 3

3 m

DB3 kN·m

A

TT

40°60°

A y

A x

5 kN4 m


A x = 1.269 kN Ans.

A y = 2.20 kN Ans.

T = 4.77 kN Ans.

5

Equilibrium equations for beam ABD4

+

+ F x = 0:

F y = 0:

M A = 0:+

A x + T cos 60° T cos 40° = 0

A y + T sin 60° +T sin 40° 5 = 0

3 kN·m + (T sin 60°)(4 m)

+ (T sin 40°)(4 m + 3 m)

(5 kN)(4 m + 3 m) = 0


25°

11. The rigid bar AB is supported by two rollers attached to its

ends at A and B. If the bar is in equilibrium in the position

shown, determine the inclination, , of the inclined plane.

20 lb

C

3 in.

7 in.

B

A


Free-body diagram of bar AB

The force acting on roller A is perpendicular

to the wall; the same is true for roller B.

Equilibrium equations for rod AB:

20 lb

25°

F B

1

2

3

F A

F x = 0:

F y = 0:

M B = 0:

+

+

+

F B cos + F

A cos 20 lb = 0 (1)

F B sin + F

A sin = 0 (2)

(F A cos )(3 in. + 7 in.) (20 lb)(7 in.) = 0 (3)

C

A

B

3 in.

7 in.


90° 25° = 65°

= 90° 65° = 25°

90°

= 90° (90° )

=

F B sin (15.447 lb) sin 25°

F B cos 20 lb (15.447 lb) cos 25°

=

tan

Solving for gives

= 47.4° Ans.

Substituting = 25° in Eq. 3 and solving gives F A = 15.447 lb.

Substituting = 25°, = , and F A = 15.447 lb in Eqs. 1 and 2 gives

F B cos + (15.447 lb) cos 25° = 20 lb (4)

F B sin + (15.447 lb) sin 25° = 0 (5)

Eqs. 4 and 5 can be solved with a calculator that can handle two

simultaneous nonlinear equations.

Alternatively, rewrite Eqs. 4 and 5 so that the constant terms are

on the right:

FB cos = 20 lb (15.447 lb) cos 25° (6)

FB sin = (15.447 lb) sin 25° (7)

Now eliminate F B by dividing Eq. 7 by Eq. 6:

6

Geometry

5

4

C

25°

A

B


A

C

B

D

12. The uniform square plate ABC of mass 10 kg is supported

by a vertical cord at B. Determine the tension in the cord and

the forces from the walls if the walls are smooth at the contact

points A and C.

20°


Free-body diagram of plate ABC

Since the wall is smooth, no friction force acts. Only

the normal force FC is present at C. Similarly only

the normal force FA is present at A.

Weight = mg = (10 kg)(9.81m/s2) = 98.1 N

Equilibrium equations for plate ABC:

F x = 0:

F y = 0:

M A = 0:

F A F

C = 0

FBD 98.1 N = 0

(FBD)(d 1) + (F

C)(d3) (98.1 N)(d2) = 0

+

+

+

(1)

(2)

(3)

1

2

3

4

A

B

C

G

F A

FC

FBD

d1

d2

d3

Center of Mass


d 1 = L sin 20°

d2A

B

G

C

45°

20°

90° 20° 45° = 25°

L cos 45°

d2 = (L cos 45°)(cos 25°)

C

B

d3

A

L2 + L2 = L 2

d3 = (L 2) sin 25°

20°

45°

20°

25°

L

L

L

L

Geometry - calculation of d3

Geometry - calculation of d 2

Geometry - calculation of d 1

Substituting these expressions for d1, d2, and

d3 into Eq. 3, canceling L, and solving Eqs. 1,

2, and 3 simultaneously gives

F A = 49.0 N Ans.

F C = 49.0 N Ans.

FBD = 98.1 N Ans.

7

8

6

5

Bd1

A

C

5.2 Rigid Bodies in Two-Dimensional Force Systems · 5.2 Rigid Bodies in Two-Dimensional Force...

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