physics 170 week 5, lecture 2 - ubc physics & astronomy |gordonws/170/2012week5/week5lec… ·...

Physics 170 Week 5, Lecture 2

http://www.phas.ubc.ca/∼gordonws/170

Physics 170 203 Week 5 Lecture 2 1

Textbook Chapter 5:Section 5.3-5.4


Learning Goals:

• Review the condition for equilibrium of a rigid body

• Illustrate the use of the condition for equilibrium of a rigidbody by solving an example.


Review: Equilibrium of a rigid body

Consider a static rigid body with a number of forces ~F1, ..., ~Fk

acting on it at points ~r1, ..., ~rk.The conditions for equilibrium are

k∑

i=1

~Fi = 0 (1)

∑

k

M◦k +k∑

i=1

~MOi =∑

k

M◦k +k∑

i=1

(~ri − ~rO)× ~Fi = 0 (2)

~M◦k are couple moments. ~MOi are moments due to forces, takenabout a point O. The sum over moments are independent of theposition of the point O. Furthermore, forces are “sliding vectors” –they can be moved along their lines of action.


Example:


The shaft assembly is supported by two smooth journal bearings A

and B and a short link DC. If a couple moment is applied to theshaft as shown, determine the components of the reaction force atthe bearings and the force in the link. The link lies in a planeparallel to the yz-plane and the bearings are properly aligned onthe shaft.


Some observations

• The reaction forces and moments at journal bearings can eachhave two components that are normal to the shaft.

• “Properly aligned” journal bearings do not have reactionmoments, only reaction forces. Therefore each journal bearingin this problem has an unknown force with two nonzerocomponents (y- and z-components). Two forces in two journalbearings accounts for four of the five unkowns in this problem.

• The link CD exerts a reaction force which is in the xz-planeand whose direction is given by the direction of the link. Themagnitude of the force in the link is the fifth unknown.


Strategy for finding a solution

• Find a mathematical description of all force vectors and thepoints of the system where they act.

• Take into account the special nature of the reaction forces andmoments at the journal bearings and at the link.

• Apply the equations for equilibrium:∑

i

~Fi = 0 , ~M◦ +∑

i

~MOi = 0 (3)

The sum of moments includes the couple moment.

• Remember that we have a choice as to where to place O.

• There is a couple moment ~M◦ = −(250Nm)i. Remember thatit is a free vector. We can place it anywhere, so we will place itat the point about which we compute the other moments.


• There are five unknowns: two components of a reaction force ateach of the journal bearings and the magnitude of the force inthe link. There are six equations of equilibrium, three are∑

i~Fi = 0 and three are ~M◦ +

∑i

~Mi=0. We can see byinspection that there are no forces in the x-direction, so oneequation

∑i Fxi = 0 is satisfied identically. The remaining five

equations are exactly the number needed to solve for the fiveunknown quantities.


Free body diagramdraw a free body diagram (not done here)


Find the coordinate vectors of the points:

~rA = (400mm)i

~rB = (−300mm)i

~rC = (120mm)i + (250mm) sin 30j + (250mm) cos 30k


Find the force vectors:

Reaction force at A:

~A = Ay j + Az k acts at ~rA = (400mm)i

Reaction force at B:

~B = By j + Bz k acts at ~rB = (−300mm)i

Force due to link is on the next page:


Force due to the link is directed along the link and has unknownmagnitude:

~FL = FL

(− cos 20j + sin 20k

)

acts at

~rC = (120mm)i + (250mm) sin 30j + (250mm) cos 30k


Compute moments:

We will take moments about the point A. Remembering that thecouple moment is a free vector, we will also place it at A. Whenmoments are taken about A, the moment due to the reaction force~A at the point A is equal to zero.

Moment due to ~B is the cross-product

~MB = (~rB − ~rA)× ~B

=(−(400mm)i− (300mm)i

)×

(By j + Bz k

)

We remembered that i× j = k and i× k = −j to get

~MB = (−700mm)(Byk −Bz j

)


Moment due to ~FL is the cross-product

~ML = (~rC − ~rA)× ~FL

=((120mm)i + (250mm) sin 30j + (250mm) cos 30k − (400mm)i

)

× FL


)

= det

i j k

−280mm (250mm) sin 30 (250mm) cos 300 −FL cos 20 FL sin 20

= i det[

(250mm) sin 30 (250mm) cos 30−FL cos 20 FL sin 20

]

− j det[−280mm (250mm) cos 30

0 FL sin 20

]

+ k det[−280mm (250mm) sin 30

0 −FL cos 20

]


Evaluating the first determinant, we get

i det[

(250mm) sin 30 (250mm) cos 30−FL cos 20 FL sin 20

]

= i(FL(250) sin 30 sin 20 + FL(250) cos 30 cos 20)

= FL(250mm) cos(30− 20)i

= FL(.250m) cos 10i

where we have used the formula

cos(a + b) = cos a cos b− sin a sin b

and we have converted mm to m.


The remining two determinants are

−j det[−280mm (250mm) cos 30

0 FL sin 20

]= FL(.280m) sin 20 j

k det[−280mm (250mm) sin 30

0 −FL cos 20

]= FL(.280m) cos 20 k

where we have also converted mm to m.

In summary, the moment due to the link is

~ML = FL(.250m) cos 10i + FL(.280m) sin 20j + FL(.280m) cos 20k


Summary of Forces:

~A = Ay j + Az k

~B = By j + Bz k

~FL = FL


)

Summary of Moments

~M◦ = −250 i Nm

~MA = 0

~MB = (−.700m)(Byk −Bz j

)

~ML = FL(.250m) cos 10i + FL(.280m) sin 20j + FL(.280m) cos 20k


Now we apply equilibrium equations: Forces must balance:

~A = Ay j + Az k

~B = By j + Bz k

~FL = FL


)

For the components:∑i Fxi = 0 automatically

∑i Fyi = 0: Ay + By − FL cos 20 = 0

∑i Fzi = 0: Az + Bz + FL sin 20 = 0


Moments must balance:

~M◦ = −250i Nm

~MA = 0

~MB = (−.700m)(Byk −Bz j

)

~ML = FL(.250m) cos 10 i + FL(.280m) sin 20 j + FL(.280m) cos 20 k

For the components:∑i Mxi = 0: − 250Nm + FL(.250m) cos 10 = 0

∑i Myi = 0: (.700m)Bz + FL(.280m) sin 20 = 0

∑i Myi = 0: (−.700m)By + FL(.280m) cos 20 = 0


Summary of linear equations:

Ay + By − FL cos 20 = 0

Az + Bz + FL sin 20 = 0

FL(.250m) cos 10 = 250Nm

(.700m)Bz + FL(.280m) sin 20 = 0

(−.700m)By + FL(.280m) cos 20 = 0

We must now solve these equations. The easiest way is by algebra.We will use this technique.


Algebraic solution of linear equations:

Ay + By − FL cos 20 = 0

Az + Bz + FL sin 20 = 0

FL(.250m) cos 10 = 250Nm

(.700m)Bz + FL(.280m) sin 20 = 0

(−.700m)By + FL(.280m) cos 20 = 0

The third equation implies

FL =250

(.250) cos 10N = 1020 N

Plug this into the other four equations to get →


Ay + By − 250N

(.250) cos 10cos 20 = 0

Az + Bz +250N

(.250) cos 10sin 20 = 0

(.700m)Bz +250N

(.250) cos 10(.280m) sin 20 = 0

(−.700m)By +250

(.250) cos 10(.280m) cos 20 = 0

Solve the third and fourth equations,

Bz = − 250N

(.250)(.700) cos 10(.280) sin 20 = −139 N

By =250N

(.250)(.700) cos 10(.280) cos 20 = 382 N


Then the first two equations imply

Ay =250N

(.250) cos 10cos 20− 250N

(.250)(.700) cos 10(.280) cos 20 = 573 N

Az = − 250N

(.250) cos 10sin 20 +

250N

(.250)(.700) cos 10(.280) sin 20 = −208

Summary:FL = 1020 N

By = 382 N Bz = −139 N

Ay = 573 N Az = −208 N


For the next lecture, please read

Textbook Chapter 5:Section 5.5-5.7


physics 170 week 5, lecture 2 - ubc physics & astronomy |gordonws/170/2012week5/week5lec… ·...

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