Equivalent forces - Practice questions for moments PDF

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Practice questions for moments
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4.5 Equivalent Force-Couple Systems

4.5 Equivalent Force-Couple Systems Example 1, page 1 of 1 1. Replace the force at A by an equivalent force and couple moment at point O. y

4 3

6m

A

2

5

MO = (16 N)(4 m) + (12 N)(6 m) = 136 N·m

4m

20 N

x

0

3 1

Calculate the moment about O.

Display the equivalent force and couple moment at point O.

Express the force in rectangular components.

(20 N) 4 = 16 N 5 A

y

A

y 6m

136 N·m

4 3

4m

5

20 N (20 N) 3 = 12 N 5

0

3 x

20 N

4 5

0

x

ns.

4.5 Equivalent Force-Couple Systems Example 2, page 1 of 1 2. The 60-N force acts at point A on the lever as shown. Replace the force at A by a force and couple moment acting at point O that will have an equivalent effect.

1

Calculate the moment about O. MO = (60 N)(200 mm) cos 60° = 6000 N·mm = 6 N·m

O

60° 200 mm

A 60 N O (200 mm) cos 60°

200 mm

60° A 60 N

2

Display the equivalent force and couple moment at point O.

O

60 N 6Nm

ns. A

4.5 Equivalent Force-Couple Systems Example 3, page 1 of 1 3. Replace the 2-lb force acting on the end of the bottle opener by an equivalent force and couple moment acting on the underside of the bottle cap at A. Use your results to explain how a bottle opener works.

A

3 in.

1

B

Calculate the couple moment about point A. MA = (2 lb)(3 in.)

2 lb

= 6 lb·in. 2

Display the equivalent force and couple moment at point A. 6 lb·in.

ns.

B 2 lb

3

The bottle opener works by pulling (with a 2-lb force) on the edge of the cap while simultaneously twisting (with a 6 lb·in. moment) the entire cap away from the top of the bottle.

ns.

4.5 Equivalent Force-Couple Systems Example 4, page 1 of 2 4. Replace the given forces and couple moment by a resultant force and couple moment at A. 260 lb

120 lb

12 13 5

800 lb·ft A

2 ft

4 ft

7 ft 2

Calculate the resultant force Rx = Fx: Rx = 100 lb

1

Express the inclined force in rectangular components.

Ry = Fy: Ry = 240 lb = 360 lb

12 (260 lb) = 240 lb 13

= 360 lb

120 lb 12 13 5

3

800 lb·ft A

Calculate the resultant couple moment about A. MAR = MA = (240 lb) (2 ft + 4 ft + 7 ft) + (120 lb) (4 ft + 7 ft) 800 lb·ft

5 (260 lb) = 100 lb 13 2 ft

20 lb

4 ft

7 ft

= 3640 lb·ft

4.5 Equivalent Force-Couple Systems Example 4, page 2 of 2 4

Determine the magnitude and direction of the resultant force. 100 lb

360 lb

5

R=

360)2 + (100)2

= 374 lb

ns.

= tan-1 360 = 74.5° 100

ns.

Display the equivalent force and couple moment at A. 374 lb

74.5° A ns.

3640 lb·ft

4.5 Equivalent Force-Couple Systems Example 5, page 1 of 3 5. Replace the force F = 3 kN acting on corner A of the block by an equivalent force and couple moment acting at the center C of the block. y F = 3 kN

A x

C

z

150 mm 150 mm 100 mm

100 mm

4.5 Equivalent Force-Couple Systems Example 5, page 2 of 3 y F = 3 kN 1 A

rCA x

C

Calculate the couple moment about C.

MC = rCA

F

= {100i

150k}mm

= 100(

i

z

j

150(

3j }kN

k

=k

j

i

j

= i

k

150 mm 150 mm 100 mm

100 mm

= { 450i

300k}kN·mm

= { 450i

300k}N m

Ans.

4.5 Equivalent Force-Couple Systems Example 5, page 3 of 3 2

Display the equivalent force and couple moment at C.

y

F = { 3j}kN MC = { 450i

300k}N·m

A x

C

z

150 mm 150 mm 100 mm

Ans.

100 mm

4.5 Equivalent Force-Couple Systems Example 6, page 1 of 4 6. Replace the forces acting on the ice auger by an equivalent force and couple moment acting at A. y

4 lb C

B 7 lb

6 in. 48 in.

A z

x

4.5 Equivalent Force-Couple Systems Example 6, page 2 of 4 y

4 lb C

B

1

Calculate the resultant force R x = F x: R x = 0

7 lb

Ry = Fy: Ry = 4 lb 6 in.

Rz = Fz: Rz = 7 lb 48 in.

A

z

x

4.5 Equivalent Force-Couple Systems Example 6, page 3 of 4 y 2

MA R = MA

4 lb C

B

Calculate the resultant couple moment about A.

= rAC

{ 4j } lb + rAB

{7k} lb

= 0, because rAC and { 4j } are parallel, or, what amounts to the same thing, the line of action of the { 4j } lb force passes throughout point A.

rAC

7 lb

rAB

{7k} lb

= 6(7)i

k

k + 48(7)j = j

48 in.

6 in.

= 0 + { 6i + 48j } in.

= {336i + 42j } lb·in.

A

z

x

i

j

=i

k

4.5 Equivalent Force-Couple Systems Example 6, page 4 of 4 y C

B

3

MAR = {336i + 42j } lb·in. A

R={

j + 7k} lb

z

x

ns.

4.5 Equivalent Force-Couple Systems Example 7, page 1 of 3 7. Replace the forces and couple moment by a single force and specify where it acts. 3 kip

A

4 kip

B

C

40°

D

20 kip·ft

2 ft

8 ft

4 ft

3 ft

4.5 Equivalent Force-Couple Systems Example 7, page 2 of 3 (4 kip) cos 40° = 3.064 kip

3 kip

y

(4 kip) sin 40° = 2.571 kip

4 kip A

B

D

C

1

20 kip ft

Resolve the inclined force into rectangular components.

40° 2 ft 2

8 ft

4 ft

3 ft

Calculate the resultant force Rx = Fx: Rx = 3.064 kip = 3.064 kip Ry = Fy: Ry = 3 kip R=

3

2.571 kip = 5.571 kip

= tan-1 5.571 = 61.2° 3.064

3.064 kip

3.064)2 + (5.571)2

= 6.358 kip

ns.

R

5.571 kip

ns.

4.5 Equivalent Force-Couple Systems Example 7, page 3 of 3 3 kip 3.064 kip

3.064 kip 4 kip 20 kip·ft

A

B

d

2.571 kip D

C

= 61.2°

40° 2 ft

8 ft

4

This is the original force-couple moment system.

6

We equate the moment of the new system, about point A, to the moment of the original force-couple system (The choice of point A for summing moments is arbitrary; any other point would work as well, except that we must use the same point for both the original system and the new system.)

C

A

5.571 kip D

4 ft

5

3 ft

7

R = 6.358 kip This is a new force-couple system that we want to make equivalent to the original force-couple system. We already know that the forces are equivalent because R is the resultant of the forces in the original system. Now we have to make sure that the moment is also equivalent. We do this by placing R at some unknown distance d from the left end and then choosing d so that the moment of this new system is the same as that of the original system.

MA R = MA or, (5.571 kip)d = 20 kip·ft

(3 kip)(2 ft)

Solving gives d = 2.10 ft

Ans.

(2.571 kip)(2 ft + 8 ft)

4.5 Equivalent Force-Couple Systems Example 8, page 1 of 5 8. Replace the forces acting on the frame by a single force and specify where its line of action intersects a) member BCD and b) member AB. y

400 N 600 N

300 N

C

B

1

D

Rx = Fx: Rx = 800 N = 800 N

2m 800 N E

A 4m

Determine the resultant force.

Ry = Fy: Ry = 400 N 2m

600 N 300 N

= 1300 N x

= 1300 N

4m 800 N

R= 1300 N

(800)2 + (1300)2 = 1526 N

= tan-1 1300 = 58.4° 800

Ans. Ans.

4.5 Equivalent Force-Couple Systems Example 8, page 2 of 5

B

800 N

58.4°

D

C

B

C

300 N

600 N

400 N

d

D 2m

1300 N R = 1526 N A

800 N

E

E

A 2

Part a) To determine where the line of action of the resultant force intersects member BCD, place the force on BCD, an unknown distance d from point B.

4m

3

2m x

4m

Choose d so that the moment about B of the resultant force equals the moment of the original force system. (The choice of point B was arbitrary.) MBR = MB or (1300 N)d = (600 N)(4 m) Solving gives d = 4.92 m

Ans.

(300 N)(4 m + 4 m)

(800 N)(2 m)

4.5 Equivalent Force-Couple Systems Example 8, page 3 of 5 B d' 800 N

D

1300 N A

C

B

300 N

600 N

400 N

58.4°

D

E

2m

R = 1526 N 800 N 4

Part b) To determine where the line of action of the resultant force intersects member AB, place the force a distance d' from point B.

A

E 4m

5

2m x

4m

Choose d' so that the moment about B of the resultant force equals the moment of the original force system.

MBR = MB or (800 N)d' = (600 N)(4 m) Solving gives d' = 8.0 m

Ans.

(300 N)(4 m + 4 m)

(800 N)(2 m)

4.5 Equivalent Force-Couple Systems Example 8, page 4 of 5 B

D

4m d' = 8 m

E

A

800 N 1300 N

58.4° R = 1526 N

6

The intersection of the line of action with a line drawn through A and B occurs below point B.

4.5 Equivalent Force-Couple Systems Example 8, page 5 of 5 7

Alternative solution for part b. Once we know where the line of action intersects member BCD, we can use geometry to find the intersection with AB.

d = 4.92 m B

C

58.4° d'

8

From triangle CBG, tan 58.4° =

d' 4.92

Solving gives, d' = 8.0 m (same result as before)

D

1300 N

A

E

G

line of action of resultant force

4.5 Equivalent Force-Couple Systems Example 9, page 1 of 4 9. Determine the value of force P such that the line of action of the resultant of the forces acting on the truss passes through the support at H. Also determine the magnitude of the resultant. 30°

160 lb

200 lb

180 lb

260 lb

B

A

D

C

P

E

8 ft F

G 6 ft

6 ft

J

I

H 6 ft

6 ft

4.5 Equivalent Force-Couple Systems Example 9, page 2 of 4 160 lb

200 lb

180 lb

(260 lb) sin 30° = 130 lb (260 lb) cos 30° = 225.2 lb

A B

D

C

P

E

30° 8 ft 260 lb G

F

6 ft

1

6 ft

Determine the resultant force. Rx = Fx: Rx = 130 lb

P

Ry = Fy: Ry = 225.2 lb

160 lb

200 lb

= 765.2 lb = 765.2 lb

I

H

(2)

180 lb

6 ft

J 6 ft

4.5 Equivalent Force-Couple Systems Example 9, page 3 of 4 160 lb

200 lb

180 lb

2

130 lb A

225.2 lb

B

D

C

P

E

8 ft G

F 6 ft

B

I

H 6 ft

J

6 ft

D

C

6 ft

3

E

Choose force P so that the moment about H of the resultant force equals the moment of the original force system about H. Any other point besides H could be used, but H has the advantage that the moment of the resultant R is zero, since R is known (as part of the statement of the problem) to pass through H.

MHR = MH : 0 = (225.2 lb)(6 ft + 6 ft) + (160 lb)(6 ft)

F

G Ry = 765.2 lb Resultant R

H

I Rx = 130 lb = 130 lb = 322.8 lb

J

(130 lb)(8 ft)

(180 lb)(6 ft) + P(8)

Solving gives

P

P = 192.8 lb

( 192.8 lb)

= 192.8 lb ns.

ns.

4.5 Equivalent Force-Couple Systems Example 9, page 4 of 4 4

Magnitude of resultant. From Eqs. 1 and 2, R=

322.8)2 + ( 765.2)2 = 831 lb

ns.

4.5 Equivalent Force-Couple Systems Example 10, page 1 of 4 10. A machine part is loaded as shown. The part is to be attached to a supporting structure by a single bolt. Determine the equation of the line defining possible positions of the bolt for which the given loading would not cause the part to rotate. Also, determine the magnitude and direction of the resultant force. y 80 N

60 N 60°

90 N·m 0.3 m

0.5 m

20 N·m

O

x 0.4 m

0.6 m

4.5 Equivalent Force-Couple Systems Example 10, page 2 of 4 y (80 N) cos 60° = 40 N 90 N·m

60 N

60° 80 N

(80 N) sin 60° = 69.28 N

1

Determine the resultant force. Rx = Fx: Rx = 40 N = 40 N Ry = Fy: Ry = 69.28 N

20 N·m

O

60 N

= 129.28 N

x

= 129.28 N 40 N

R=

129.28 N

40)2 + (129.28)2 = 135.33 N

= tan-1 129.28 = 72.8° 40

Ans. Ans.

4.5 Equivalent Force-Couple Systems Example 10, page 3 of 4 y

y

60 N

40 N

90 N·m 60° 80 N

40 N (x, y)

0.3 m 72.8°

69.28 N

O

2

O

x 0.4 m

0.6 m

Rotation is caused by moment. If the given force-couple system is replaced by a single equivalent force (the resultant) and a specified line of action, then the moment would be zero about any point on the line of action. So the line of action is the line on which the bolt should be placed to prevent rotation.

129.28 N

y

0.5 m

20 N·m

R

x x

3

To find the equation of the line of action, place the resultant at a general point (x, y). Then equate moments about, say, point O for the resultant (the figure on the right) and the original loading (the figure on the left): MOR = MO or (40 N)y (129.28 N)x = 90 N·m 20 N·m + (40 N)(0.3 m + 0.5 m) (69.28 N)(0.4 m) (60 N)(0.4 m + 0.6 m)

4.5 Equivalent Force-Couple Systems Example 10, page 4 of 4 4

Solving for y gives the equation of a line y = 3.232 x + 0.357

5

All points at the top of the machine part have a y coordinate of y = 0.3 m + 0.5 m

Ans.

This line defines the possible locations of the bolt.

= 0.8 m. Substituting y = 0.8 m into the equation of the line for the bolt locations, y = 3.32 x + 0.357

0.133 m

and solving for x gives x = 0.133 m

y

0.3 m

0.5 m 0.357 m 0

x 0.4 m

0.6 m

4.5 Equivalent Force-Couple Systems Example 11, page 1 of 4 11. The rectangular foundation mat supports the four column loads shown. Determine the magnitude, direction, and point of application of a single force that would be equivalent to the given system of forces. y 30 kip 15 kip A 20 kip

11 kip

B z C

10 ft

D 3 ft

5 ft

x

8 ft 1

Determine the resultant. Ry = Fy: Ry = 30 kip = 76 kip = 76 kip

15 kip

20 kip

11 kip

4.5 Equivalent Force-Couple Systems Example 11, page 2 of 4 2 To make the resultant force R equivalent to the original system of forces, place R at the point (x, 0, z) and then determine values of x and z such that R z produces the same moments about the x and z axes as the given forces produce.

y 30 kip 15 kip

A 11 kip

20 kip B z

D

C

10 ft

8 ft

3

20 kip

x, A

D

C, B 8 ft

x

z (x, 0, z)

D

C

x

First, equate moments about the x axis. We can use either the scalar definition of moment, M = Fd, or the vector product definition. Let's use the scalar definition. R = 76 kip

y

30 kip

15 kip z

B

y 11 kip

A

R = 76 kip

x

3 ft 5 ft

y

z

D

C, B

3 ft

x, A

z View from positive x axis

4

MxR = Fd: (76 kip)z = (20 kip)(8 ft + 3 ft) + (15 kip)(8 ft + 3 ft) + (11 kip)(3 ft ) + (30 kip)(0)

(1)

4.5 Equivalent Force-Couple Systems Example 11, page 3 of 4 5 Solving Eq. 1 gives

y

z = 5.5 ft

30 kip

y Ans.

A

R = 76 kip

Next, equate moments about the z axis.

15 kip A 20 kip

B

11 kip

z

B

x

z (x, 0, z)

D

C

z C

10 ft

x

D 3 ft

5 ft

x

8 ft y

6 Use two-dimensional views.

30 kip 15 kip

z, B, A 10 ft

7

M zR = Fd:

20 kip

11 kip

C

D 5 ft

y

R = 76 kip C

z, B,...