Static Machine Forces

Determines the forces in a machine while links are stationary, or at constant velocity.

Often used when the acceleration of the machine components is small (<0.25g).

Section 1

Forces

Push or pull action. Vector quantity.

– Magnitude– Direction– Point of application

F = 100 lbs 300

Force Components

It is often convenient to resolve a force into orthogonal components.

Use right triangle to determine the components– F1=F sin– F2=F cos

F

F1F2

1

2

Moments

Twisting action produced by a force. Computed relative to a point. MA=F d

– F = Force– A = Reference point– d = perpendicular distance between force and

reference point.

Vector

Fd

A

MA = 250 in lbs, cw

Problem 1-2

A force is applied to a box wrench as shown. Determine the moment, relative to the center of the nut, when = 300, and = 700.

25 lb

8 in.

Problem 1-4

A force is applied to the control lever as shown. Determine the moment, relative to the pivot block, when = 300.

60 lb

18 in

12 in

Free-Body Diagrams

Isolate the component(s) that must be studied.

Draw the component as if it were floating freely.

Replace all supports with the appropriate force and/or couples (moments).

Free-Body Diagrams

C

AB

D EF

FBDEntire engine hoist

C

AyBy

D EF

250 lbs

Bx

FBDLink DEF

CE

D

Dy

EF

250 lbs

Dx

By

FBDBase

C

Ay

D

Bx

Dy

Dx

CE

FBDCylinder CE

CE

CE

Free-Body Diagrams

Reaction Forces

As a general rule, if the nature of the contact prevents motion in a certain direction, there must be a supporting force in that direction.

F

Two-Force Members

A component that is acted upon by only two forces is known as a two-force member.

A two force member will always be in either tension or compression.

Problem 1-12

Draw free-body diagrams of the links for the pliers when 10 lbs is applied to the handles

3.75 in1.25 in 10 lbs

Problem 1-13

Draw free body diagrams of all the components for the lift. The weight of the crate and platform are 1200 lb and 400 lbs. The weight of all other links is considered insignificant

18”

18”

6”

24” 24” 24”

80”

10”

Static Equilibrium

A machine component is in static equilibrium when the combination of all forces is zero.

In addition, the net effect of all moments, about any arbitrary point, must also result in zero.

Fx = 0 Fy = 0 i = 0

Hints to Solve Statics Problems

Draw free-body diagrams large and properly labeled all forces.

Identify any two-force members that are contained in the machine.

The shortest method of solution occurs when the first FBD incorporates both given information and what is required. However, a free body diagram can, at most, be used to determine three unknown forces.

Hints (con’t) If the first FBD fails to give a full solution, draw

several other free-body diagrams. Each additional FBD generates three additional independent equations. However it may generate additional unknown forces. Keep track of the total number of unknown forces and independent equations.

A FBD of the entire machine may be useful. Remember that internal forces between

connecting components are equal and opposite.

Problem 1-12

Determine the force onto the nut when 10 lbs is applied to the handles of the pliers.

3.75 in1.25 in 10 lbs

Problem 1-14

The clamp has a rated load of 1500 lb. Determine the compressive force this creates in the threaded rod, AB

5 in2 in

5 in

4 in

1500 lb3 in

2 in

Problem 1-13

Determine the cylinder force to maintain the position of the lift. The weight of the crate and platform are 1200 lb and 400 lbs. The weight of all other links is considered insignificant

18”

18”

6”

24” 24” 24”

80”

10”

45”

Problem 1-25

Determine the force from the hydraulic cylinder to keep the platform in the position shown. 500 lb24”

36” 36”

28”

28”

30”

10”

20.6”

12.9”

37.90

37.90

75.60

16”

Problem 1-16

Determine the force required by the hydraulic cylinder to maintain position of the bucket.

1.2 m

2.4 m

5 m

0.9 m2.0 m1200 N