Chapter 6

Concurrent and Parallel Forces

Introduction

• Forces do not always cause motion– Static = not moving: may be in

equilibrium– Example: Bridge

• Forces acting nonconcurrently– May produce rotational motion– Torque

Objectives

• Vector sum of concurrent forces

• Equilibrium in one dimension• Force diagrams• Compression and tension• Torque & rotational

problems• Parallel force problems• Conditions for equilibrium• Center of gravity

6.1 Forces in Two Dimensions

• Concurrent forces: forces applied to or act at the same point

• Resultant force: sum of forces applied at same point– Single force that has same effect as

the two or more forces acting together– When acting in one dimension, add

and subtract based on direction– In 2 dimensions, use vectors

Finding Resultant Vectors• Use x and y-component

method• Use Vector Triangle

Method• Use Parallelogram Method

x and y-component method

• Forces F1 and F2 act concurrently (at same point)

• Find resultant force

F1 = 50 N @ 35°

F2 = 60 N @ 70°

x and y-component method

• Find x and y-components using right angle trigonometry

• Add x and y-components to get resultant force F2 = 60 N @ 70°

F2x = 60 cos 70°

F2y = 60 sin 70°

F1x = 50 cos 35°F1y = 50 sin 35°

F1= 50 N @ 35°

Vector Triangle Method

• Put vectors to be added end to end

• Resultant vector is side of triangle

R = F1 + F2

Parallelogram Method

R = F1 + F2

See examples in textbook on pages 153 to 155

6.2 Concurrent Forces in Equilibrium

• Equilibrium is the state of a body in which there is no change in motion

• Net force acting on body is zero

• No acceleration: at rest or moving at constant velocity

• Study of objects in equilibrium is called statics

Equation for Equilibrium (one dimension)

• F+ = F-

– F+ = sum of forces acting in one direction (positive)

– F- = sum of forces acting in opposite (negative) direction

– +/- can be up & down or left & right, etc.

• See examples on page 158

Conditions for Equilibrium (2 dimensions)

• Equilibrant force: when two or more forces act at one point, this is the force that when applied at the same point produces equilibrium

• Is equal in magnitude to resultant force, but opposite in direction

Equilibrium (2-D) continued• If object is in equilibrium in

two dimensions, the net force acting on it must be zero

• Sum of x-components = 0• Sum of y-components = 0• Both must be true for

equilibrium• See examples on pp. 160-163

Solving Equilibrium Problems

• Draw force diagram from point at which unknown forces act

• Find x and y-components of each force

• Substitute components into equations

• Solve for unknowns

Tension

• Stretching force produced by forces pulling outward on the ends of an object

• Example: rubber band

Compression

• Force produced by pushing inward on the ends of and object

• Example: valve spring

6.3 Torque

• Torque is the tendency to produce change in rotational motion

• Torque is produced when a force is applied to produce rotational motion (ex: using wrench to turn a bolt)

Torque: Equation

• Depends on two factors– Amount of force applied– Distance from point of

rotation that force is applied

• τ = Fst

– τ = torque (N∙m or lb∙ft)– F = applied force (N or lb)

– st = length of torque arm (m or ft)

Be sure to use perpendicular distance!

• See bicycle example on pg 167

• Measure distance that is perpendicular to applied force

Right-hand Rule

• Torque is vector quantity that acts along axis of rotation

• Use right hand rule to determine direction– Grasp axis of rotation with

right hand so that fingers circle in direction of rotation

– Thumb points in direction of torque vector

6.4 Parallel Force Problems• Solve using conditions of

equilibrium• First condition of equilibrium: the

sum of all parallel forces on a body in equilibrium must be zero (ΣF = 0)

• Second condition of equilibrium: the sum of the clockwise torques on a body in equilibrium must equal the sum of the counterclockwise torques about any point (Στcw = Στccw)

To solve parallel force problems (pp 169-173)• Sketch the problem• Write equation setting sums of

opposite forces equal to each other• Choose point of rotation (eliminate

variable if possible)• Write sum of clockwise torques• Write sum of counterclockwise

torques• Set Στcw = Στccw (solve for

unknown)• Substitute into force equation to

solve for unknown

6.5 Center of Gravity

• The center of gravity of any object is the point at which all of its weight can be considered to be concentrated

• In uniform material, will be geometric center of object

• Many objects, weight is not evenly distributed (ex: car, person)

Irregular Object: Center of Gravity

• Point at which object will balance

• Weight represented as vector through center of gravity of the object

• See examples pp 175-176• Try activity on page 176

Problems

• 6.1 (pp 155-157) #4, 6, 12• 6.2 (pp 164-166) #2, 4, 10,

12, 20, 24• 6.3 (pp 168-169) #2, 8, 16• 6.4 (pp 173-174) #4, 8, 10,

14• 6.5 (pp 177-178) #4, 8, 14