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STATIC EQUILIBRIUM
PROVINCIAL EXAMINATION ASSIGNMENT
ANSWER KEY / SCORING GUIDE
PART A: Multiple Choice (each question worth ONE mark)
VERSION 1.0
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-1-
1.
A 25 kg droid rests on a 5. 0 m long shelf supported by two cables as shown. The mass of
the shelf is 12 kg.
droid
Q2D2
0.60 m
0.60 m
0.80 m
3.8 m
Find the tension in each cable.
(7 marks)
Using left-hand support as fulcrum:
∑ τc
= ∑ τ cc
τ D + τs = τc
FD d D + Fs ds = Fc R dc
25 kg ⋅ 9.8 m s2 ⋅ 0.80 m + 12 kg ⋅ 9.8 m s2 ⋅1. 9 m = Fc R ⋅ 3.8 m
Fc R = 110 N
Fc L + Fc R = Fg
986phk



 ← 1 mark



← 3 12 marks
←
1
2
mark
← 1 mark
Fc L + 110 N = 363 N
←
1
2
mark
Fc L = 253 N
←
1
2
mark
-2-
July 24, 1998
2.
Peter exerts a horizontal force F on a 12 kg bucket of concrete so that the supporting rope
makes an angle of 20° with the vertical.
20°
F
a)
Find the tension force in the supporting rope.
FT
Fp
FT
(3 marks)
20°
Fg = 118 N
← 1 marks
Fg
Fp
FT =
118
cos 20°
←1
= 125 N
marks
← 1 mark
FT = 1.3 × 10 2 N
b)
Peter now exerts a new force which causes the rope to make a greater angle with the
vertical. How will the tension force in the supporting rope change?
4
p
p
p
The tension force will increase.
The tension force will decrease.
The tension force will remain the same.
(Check one response.)
988phpk
(1 mark)
-3-
September 8, 1998
c)
Using principles of physics, explain your answer to b).
(3 marks)
The vertical component of the tension is equal to the weight and is unchanged. Peter’s
horizontal force increases with a larger angle. The horizontal component of the tension is
equal to Peter’s and therefore is also increased. Thus, the resultant tension is increased.
988phpk
-4-
September 8, 1998
3.
A uniform 4.8 m long ladder of mass 16 kg leans against a frictionless vertical wall as shown in
the diagram below.
4.8 m
65°
a)
Draw and label a free body diagram showing the forces acting on the ladder.
(2 marks)
Fw
Fg
2 marks
Fy
Fx
b)
What minimum force of friction is needed at the base of the ladder to keep it from sliding?
(5 marks)
τ cw = τ ccw
Fg (2.4) = Fw ( 4.8)
⊥
⊥
← 3 marks
(16)(9.8)(cos 65)(2.4) = Fw sin 65( 4.8)
159 = Fw ( 4.35)
Fw = 37 N
∑F
x
= 0 Fx = Fw
Ff = 37 N
968phpk
← 1 mark
← 1 mark
-5-
December 13, 1996
4.
A uniform beam 6.0 m long, and with a mass of 75 kg, is hinged at A. The supporting cable keeps
the beam horizontal.
cable
A
37˚
load
2.0 m
3.5 m
6.0 m
If the maximum tension the cable can withstand is 2.4 × 103 N , what is the maximum mass
of the load?
(7 marks)
τCW = τCCW
(
← 1 mark
)
F⊥ = 2. 4 × 103 sin 37° 
 ← 2 marks

= 1 444.3 N
∴ Using torque about A:
3. 0 ( 735 ) + 3. 5( F L ) = 1 444.3( 2. 0 ) ← 3 marks
2 205 + 3. 5( F L ) = 2 888. 6 N 


3. 5( F L ) = 683 N

Load = 195. 4 N 

 1 mark
FL

Mass =
9.8

= 19. 9 kg 


= 20 kg
946phpk
-6-
June 11, 1996
5.
A 6. 0 m uniform beam of mass 25 kg is suspended by a cable as shown. An 85 kg object
hangs from one end.
Cable
Beam
(25 kg, 6.0 m)
67°
75°
4.0 m
85 kg
What is the tension in the cable?
(7 marks)
T
67°
∆ 75°
( 25 kg )( 9. 8 m s
2
75°
) = 245 N
(
833 N = ( 85 kg ) 9. 8 m s
2
)
Στ = 0
τ c = τ cc
τ 245 + τ 833 = τ T
3. 0 m ( 245 N ) sin 75° + 6. 0 m ( 833 N ) sin 75° = 4. 0 m T sin 67°
← 1 mark
← 5 marks
710 N ⋅ m + 4 830 N ⋅ m = 3. 68 m T
5 540 N ⋅ m = 3. 68 m T
1 500 N = T
991phk
-7-
← 1 mark
February 23, 1999
6.
The diagram shows the rear door of a station wagon supported horizontally by a strut. The mass of
the door is 18 kg and the compression force in the strut is 450 N.
centre of gravity
x
0.36 m
32 °
F = 450 N
strut
a)
Draw and label a free body diagram showing the forces acting on the door.
(2 marks)
Fs [ 12 mark]
F [1 mark]
b)
Fg
[ 12 mark]
At what distance, x, from the hinge is the centre of gravity of the door located?
(5 marks)
τ c = τcc
mgx = Fs ( d )sinθ
← 2 marks
18( 9.8 ) x = 450( 0.36 )sin 32° ← 2 marks
x = 0.49 m
976phpk
← 1 mark
-8-
August 14, 1997
7.
A 0.75 kg board of length 2.60 m initially rests on two supports as shown.
0.40 m
a)
1.40 m
x
What maximum distance, x, from the right-hand support can a 1.20 kg bird walk before the
board begins to leave the left-hand support?
(5 marks)
Take torques about right support
τC = τ C C
b)
← 1 mark
1. 20 ( 9.8 ) x = 0. 75( 9.8 ) ( 0. 50 )
123
1 mark
← 2 marks
x = 0.31 m
← 1 mark
What force does the right-hand support exert on the board at that instant?
F up = F down
1
2
mark
F = 1. 2 ( 9.8 ) + 0. 75( 9.8 )
1
2
mark
F = 11. 76 + 7.35
1
2
mark
1
2
mark
= 19 N
956phpk
- 9-
(2 marks)
May 1, 1996
8.
An object of mass, m, is suspended by two cords connected to a wall and to a 5.0 kg block resting
on a table as shown.
32 °
5.0 kg
µ = 0. 47
m
A coefficient of friction of 0.47 exists between the 5.0 kg block and the table. What is the
maximum mass, m, that can be hung from the cords before the 5.0 kg block begins to move?
(7 marks)
FT
≡
F
Fg
32 °
FT
Fg
F
F f = µFN
= 0. 47 × 5. 0 × 9.8
= 23 N
F
tan 32°
F
mg =
tan 32°
F
m=
g × tan 32°
23
m=
9.80 × tan32°
Fg =
m = 3.8 kg
0101phk
← 2 marks




 ← 4 marks




← 1 mark
- 10 -
February 23, 2001
9. A 6. 0 m uniform beam of mass 32 kg is suspended horizontally by a hinged end and a
cable. A 93 kg object is connected to one end of the beam.
Cable
6.0 m beam, 32 kg
48 °
R
4.0 m
Hinge
93 kg object
What is the magnitude and direction of the reaction force R that the hinge exerts on the beam? (10 marks)
τ c = τ cc about the hinge 

3. 0 ( 314 ) + 6. 0 ( 911) = 4. 0 ( T ) sin 48°

 ← 2 mar ks
942 + 5 470 = 2. 97 T


2 160 N = T



T cos 48 = Rx

 ← 1 mar k
2 160 cos 48 = Rx


3
Rx = 1 445 N (1. 45 ´ 10 N)
T x = Rx
S t = 0 about the hinge 

( 314 )( 3. 0 ) - T y ( 4. 0 ) + 911( 6. 0 ) = 0
 ← 2 mar ks

T y = 1 600 N

R y = Ty - WLoad - WBeam
ü
ï
= 1600 N - 911N - 314 N ý ¬ 2 marks
ï
= 375 N
þ
998phk
- 11 -
Tx
T
Ty
48°
q
Ry
R
314 N
¬ 1 mark
911 N
Rx
üï
ý ¬ 1 mark
= 1491 N = 1.49 ´103 N ïþ
R = Rx2 + R y2
ü
æRy ö
÷÷
q = tan -1 çç
ï
ý ¬ 1 mark
èRx ø
ï
0
q = 15 below horiz .þ
September 8, 1999
10. A 65 kg person is 43 of the way up a 25 kg uniform ladder as shown in the diagram below. The ladder
is leaning against a frictionless surface inclined at 600 to the horizontal. What is the minimum coefficient
of friction between the ladder and the floor necessary to maintain equilibrium? (10 marks)
FNW
300
400
200
Identify all angles } 1 mark
FNF
WP
FF
åt
ccw
400
WL
1200
600
= å tcw
1
3
ü
FNW (sin 70 )d = WL (sin 50 ) d + WP (sin 50 ) d ï
2
4
ï
1
3 ï
FNW (sin 70 )d = (245)(sin 50 ) d + (637 )(sin 50 ) d ý 3 marks
2
4 ï
FNW = 489 N
ï
ï
þ
åF
=0
åF
=0
x
FF = FNW (cos 30 ) ü
ï
= (489 )(cos 30 )ý 1 mark
ï
= 423N
þ
y
WL = (25)(9.8) = 245 N ü
ý 1 mark
WP = (65)(9.8) = 637 N þ
FNF = Wl + WP - FNW (sin 30 )
ü
ï
= 245 N + 637 N - (489 )(sin 30 )ý 2 marks
ï
= 638 N
þ
FF = mFNF
ü
ï
FF
423
ý 2 marks
\m =
=
= 0.66ï
FNF 637
þ
- 12 -
11.
A wire is stretched between two posts. A mass is suspended near the centre as shown below.
If the tension in the wire were increased, is it possible to make the wire perfectly horizontal?
Explain your answer in terms of forces.
(4 marks)
No, it is not possible to make the wire perfectly horizontal. Since the mass has a vertical force of
gravity acting on it, the tension in the wire must have an opposite vertical component. A horizontal
tension has no vertical component; therefore, it is not possible to make the wire perfectly horizontal.
968phpk
- 13 -
December 13, 1996