Physics 170 Lecture 10 Chapter 4 - “Force System ...mattison/Courses/Phys170/p170-10.pdf ·...

Phys 170 Lecture 10 1

Physics 170 Lecture 10

Chapter 4 - “Force System Resultants”

A wrenching experience.....


M = 5i + 6 j + 7k( ) ft-lb

PROBLEM 4-116 (page 169 12e)

The pipe assembly is acted on by forces

!F1 and

!F2 and by a couple

moment

!M = (5

!i + 6

!j + 7

!k) lb !ft .

• Determine the magnitude and coordinate direction angles of !M .

• Determine the magnitude of each of the two forces comprising the

couple when the lever arm of the couple is 0.5 ft.

• Replace the force-couple system by a resultant force and couple

moment at O. Express the results in Cartesian vector form.

y


M = 5i + 6 j + 7k( ) ft-lbM = 52 + 62 + 72 = 10.49 ft-lb

α = cos−1 510.49

= 61.5° β = cos−1 610.49

= 55.1°

γ = cos−1 710.49

= 48.1°



!F1 and

!F2 and by a couple

moment

!M = (5

!i + 6

!j + 7

!k) lb !ft .








!F1 and

!F2 and by a couple

moment

!M = (5

!i + 6

!j + 7

!k) lb !ft .






M = 10.48 ft-lb = Fd = F ⋅0.5 → F = 20.97 lb


M = 5i + 6 j + 7k( ) ft-lb



!F1 and

!F2 and by a couple

moment

!M = (5

!i + 6

!j + 7

!k) lb !ft .






y


FR =

F1 +F2 = −20, −10, 25( ) + −10, 25, 20( )

= −30i +15 j + 45k( ) lb

R1 = 1.5, 2, 0( )

F1 = −20, −10, 25( )

M1 =

R1 ×

F1 =

i j k1.5 2 0−20 −10 25

= 50, −37.5, 25( )

R2 = 1.5, 4, 2( )

F2 = −10, 25, 20( )

M 2 =

R2 ×

F2 =

i j k1.5 4 2−10 25 20

= 30, −50, 77.5( )

MR =

M1 +

M 2 +

M = 85i − 81.5 j +110k( ) ft-lb


M = 5i + 6 j + 7k( ) ft-lb



!F1 and

!F2 and by a couple

moment

!M = (5

!i + 6

!j + 7

!k) lb !ft .






y

MR = 85i − 81.5 j +110k( ) ft-lb

FR = −30i +15 j + 45k( ) lb


PROBLEM 4-137 (page 182, 13th

edition)


edition)

• Replace the three forces acting on the plate by a wrench.

• Determine the magnitude of the force and couple moment of the

wrench, and the point P(x ,y,0) where its line of action intersects

the plate.

“Simplifying” Force Systems

Say we have n forces and m moments about some origin.

We can always add the forces into one “resultant force” FR

and add the moments into one “resultant moment” MR.

That “simplifies” the problem from n+m vectors to 2 vectors.

We can pretend FR and MR are both applied at the origin.


“Simplifying” Force Systems (2)

If all the forces are parallel, MR has to be perpendicular FR.

In that case, it is possible to find a place where applying FR would itself create MR.

You already know an example of this: the Center of Gravity.Gravity is a distributed force, but the total force, and also the total moment from it, acts like it’s applied at the CoG.



Moving FR from the origin can only generate a moment that is perpendicular to FR, not parallel to it.

If the forces aren’t parallel, MR can point in any direction, so it can have a component parallel to FR, which we can’t generate by moving FR.

But, we can always break MR

into a component parallel to FR, and a component perpendicular to FR.



And moving FR from the origin can generate the componentof MR that is perpendicular to FR.

And it never matters where a pure couple-moment like M|| is applied. So we can move it to the same place we moved FR.

Reduction to a wrench: FR at a special point, and M|| Phys 170 Lecture 10 11




edition)


edition)

• Replace the three forces acting on the plate by a wrench.

• Determine the magnitude of the force and couple moment of the

wrench, and the point P(x ,y,0) where its line of action intersects

the plate.

FR =

FA +

FB +

FC = 500i + 300 j + 800k( ) N

Make life easier by putting origin at A ! →MA = 0

MB = 4 ⋅800i by right-hand ruleMC = 6 ⋅300k by right-hand ruleMR = 3200i + 0 j +1800k( )Mwrench = uFR ⋅

MR = 3070.86 N-m

Mwrench =

Mwrench uFR

= 1551i + 930.6 j + 2481k( ) N-m

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