october 13 2010 midterm exam questions and answers

8/3/2019 October 13 2010 Midterm Exam Questions and Answers

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THE UNIVERSITY OF BRITISH COLUMBIA

2010/2011 WINTER SESSION

MIDTERM EXAMINATION

PHYSICS 170 MECHANICS I

2:00 pm to 2:50 pm, Wednesday, October 13, 2010

MARKS

This exam counts for a total of 20 marks towards your Final Grade for the course.

The exam consists of 2 questions each worth 20 marks.

Write all work to be marked in the Answer Booklets provided.

NUMERICAL ANSWERS

You must give numerical answers correctly to three figures and with correct units to get any marks for your

numerical answers. No marks will be given for correctly solving incorrect equations.

CALCULATOR

You may use a graphing calculator.

INFORMATION SHEET

You may consult one 8 ! inch by 11 inch hand-written double-sided Information Sheet.

Your Information Sheet must not contain any sample problems or solutions to sample problems.

Put your name and student number on your Information Sheet. You must hand in your Information Sheet

with your Answer Booklet in order for your exam to be marked.


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PHYSICS170MECHANICS Midterm Exam: 2:00pm to2:30 pm, Wednesday, ctober13,2010OUESTION (20marks)

Thediagrambelowshowshree cablesattache!to pointl on the z-axis and o pointsB, C andD on the rry-planeA force F of magnitude 000N acts n the z-directionat pointu4. Point is in equilibrium.

a) Draw a large,clear ree-bodydiagram or point,4 showing he force F and he forces FAo F^,exertedonI by thecables. Showalsothe Cartesian nit vectorsand he numericalvaluesof thecoordinates f A, B, C andD. ;

b) ExpressF*, Fo, and F* asCartesianveplors.c) Detennine he Cartesian omponeNrtequationsf equilibrium or pointl.d) Solve heCartesian omponent quationsof equilibrium for pointI to determinehe magnitudes

of F*, Fo" andF*.

and FA

(4 mar(6 mar(6 mark

(4 markf!*II{

$ fil


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ANSWERS TO QUESTION 1

b) Forces!

FAB

= (2!

i +3!

j!6!

k)X X= FAB

/ 22+3

2+6

2

!

FAC = (!1.5!

i +2!

j!6!

k)Y Y = FAC / 1.52+22+62

!

FAD

= (!3!

i !6!

j!6!

k)Z Z= FAD

/ 32+6

2+6

2

!

F= 1000!

k

c) Cartesian component equations of equilibrium

! Fx= 0: 2X!1.5Y! 3Z= 0

! Fy= 0: 3X+2Y!6Z= 0

! Fz= 0: !6X!6Y!6Z+1000 = 0

d) Solution to the Cartesian component equations of equilibrium

(X,Y, Z) = (94.34, 18.87, 53.46)

(FAB

, FAC

, FAD

) = (660, 123, 481) N


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PHYSICS170MECHAIIICS I Midterm Exarn: 2:00pm to 2:50pm, Wedrtesday, ctober13,2010OUESTION2 (20marks)

The diagrambelow showsa rectangular late n thery-plane. Theplate s actedonby forces F* Fr and d asindicated nthediagram. Fr^uybe aken o actat (2, 6, 0) m.

I

a) Draw a large,clear diagram or theplateshowing he forces F* F, and 4 .vectorsand henumericalvaluesof the coordinateswhere he forcesact.

b) Express4 *d { as Cartesiarvectors. :i\c) Determine heresultantmoment frI* of F, and F,

d) Determine he magnitud of fr * .e) Determine he coordinate irectiorrangledof fi* .

F : * 4 k N

Showalso heCartesian n

aboutpoint O.

(3 mark(4 mark(6 mark(3 mark(4 mark


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ANSWERS TO QUESTION 2

b) Forces!

F2= F

2x

!

i +F2y

!

j+F2z

!

k

F2x= !5cos45!sin30! F

2y= 5cos45!cos30! F

2z= 5sin45!

!

F2= (!1.77

!

i +3.06!

j+3.54!

k) kN

!

F3= F

3x

!

i +F3y

!

j+F3z

!

k

F3x= 4cos60! F

3y= 4cos60! F

3z= 4cos45!

!

F3 = (2.00

!

i +2.00

!

j+2.83

!

k) kN

c) Resultant moment

(!

MR)O= M

x

!

i +My

!

j+Mz

!

k= !!

r "!

F

=

!

i!

j!

k

2 6 0

0 0 6

+

!

i!

j!

k

4 0 0F

2x F2y F2z= 36

!

i!(12+ 4F2z)

!

j+ 4F2y

!

k

= (36.0!

i !26.1!

j+12.2!

k) kN m

d) Magnitude of the resultant moment

MRO = M2

x+My2

+Mz2

= 46.1 kN m

f) Coordinate direction angles of the resultant moment

! = cos"1(Mx/M

RO

)= 38.7! != cos"1(My/M

RO)=125!

!= cos"1(Mz/M

RO

)= 74.6!

october 13 2010 midterm exam questions and answers

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