3rd Class - Edition 2.0
Solutions to Self Test Problems
A1 - CHAPTER 4
1. A boat is traveling on a bearing of 25° east of north at a speed of 5 knots (a knot is 1.852 km/h). A er
traveling for 3 hours, the boat heading is changed to due south and it travels for a further 2 hours at 5
knots. What is the boat’s bearing from its original posi on?
Since the speed doesn’t change, we can draw a vector diagram using me and direc on.
First, draw a scaled line represen ng 3 hours at 25° E of N.
Then, draw a scaled line represen ng 2 hours at due south.
Finally, draw a line from the ending point to the origin and measure the angle.
Resultant bearing measured at 60.4° E of N. (Ans.)
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2. Find the magnitude and direc on of the resultant of 4 concurrent coplanar forces of:
100 N ac ng 50° west of north
40 N ac ng west
60 N ac ng 20° east of south
15 N ac ng east
Using an appropriate scale (eg., 1 cm = 10 N):
• Draw each force vector in its required direc on, star ng from the origin with the first force.
• The resultant is the line drawn from the origin to the end of the last vector drawn.
• Measure the length and direc on of the resultant.
With allowances for drawing accuracy,
the resultant in this case should be: 81-82 N @ about 84° W of N (Ans.)
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3. Find the resultant and the equilibrant for the following system of concurrent forces.
150 N, 60° north of west
100 N, 45° west of south
200 N, 20° south of east
The scaled used is: 1 cm = 20 N
Draw the 3 force vectors in order as shown.
The resultant, from the origin to the end of the last force vector is measured as:
2.2 cm at 13° S of E
 The resultant is 2.2 x 20 = 44 N at 13° S of E (Ans.)
The equilibrant is directly opposing the resultant, so is 44 N at 13° N of W (Ans.)
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4. If the following forces are applied to a body, what addi onal force is required to keep the body in
balance:
50 N ac
30 N ac
60 N ac
15 N ac
ng 40° east of north
ng east
ng 20° north of west
ng south
A er drawing the four force vectors as shown, the resultant (from the origin to the end of the 15 N
force vector) is measured as:
4.4 cm at 8° E of N
 The resultant force is: 4.4 x 10 = 44 N at 8° E of N
The force required to balance this system is the equilibrant which is directly opposing the resultant.
 The required force is: 44 N at 8° W of S (Ans.)
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5. A force of 120 N is applied at an angle of 47° to the south of west. What are the ‘x’ and ‘y’ components of this force?

- x
cos 47° =   _____
    
120 N
x = - 120 x cos 47°
= - 120 x 0.6819
= - 81.8 N (Ans.)

- y
    
sin 47° =   _____
120 N
y = - 120 x sin 47°
= - 120 x 0.7313
= - 87.8 N (Ans.)
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6. The components of a certain force are 95 N east and 135 N south. Describe the force.

First use Pythagarus to find magnitude.
2
+ 1352
F2 = 95
_________
F = √______
952 + 1352  
= √ 27 250  
= 165.1 N
Now use trigonometry to find direc on.
135
tan = ____
    = 1.4211
95
  = tan−1 1.4211
= 54.8°
 The force is described as: 165.1 N ac ng at 54.8° S of E (Ans.)
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7. A force system consists of these three forces: 400 N at 30° north of east, 250 N at 40° north of west,
and 200 N at 60° south of east. Using components, find the equilibrant for this system. Check your
calculated answer, using the graphical method (that is, by drawing vector diagram).
Consider the 400 N force:
x component = 400 cos 30°
= 400 x 0.866 = 346.4 N (+)
y component = 400 sin 30°
= 400 x 0.5 = 200 N (+)
Consider the 250 N force:
x component = −250 cos 40°
= −250 x 0.766 = −191.5 N
y component = 250 sin 40°
= 250 x 0.6428 = 160.7 N (+)
Consider the 200 N force:
x component = 200 cos 60°
= 200 x 0.5 = 100 N (+)
y component = −200 sin 60°
= −200 x 0.866 = −173.2 N
Sum of x components = 346.4 − 191.5 + 100 N
= + 254.9 N
Sum of y components = 200 + 160.7 − 173.2 N
= + 187.5 N
(Continued)
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7. (Continued)
To find the resultant:
2
+ 187.52
R2 = 254.9
______________
R = √ 64 974 + 
35 156  
R = 316.6 N
187.5
sin  = _____
 
  = 0.5922
316.6
 = sin−1 0.5922 = 36.3° N of E
The equilibrant is directly opposing the resultant, so is:
316.6 N at 36.3° S of W (Ans.)
Check graphically by drawing the forces to scale, then measuring the angle and resultant:
Scale: 1 cm = 100 N
R = 3.15 cm
= 315 N
 = 37°
This agrees, with slight errors due to drawing accuracy.
• 3rd Class - Edition 2.0
8. Four forces act on a single point. The forces are defined as: 60 N due north, 125 N at 35° east of
south, 75 N at 62° south of east, and 80 N at 20° south of west.
a) Find the total horizontal components of the forces.
horizontal component of ‘n’ of 60 N force = 0 N
horizontal component of 75 N force = +75 cos 62°
= +75 x 0.4695
= +35.21 N
horizontal component of 125 N force = +125 cos 55°
= +125 x 0.5736
= +71.7 N
horizontal component of 80 N force = −80 cos 20°
= −80 x 0.9397
= −75.18 N
  
 Total horizontal components = 0 + 35.21 + 71.7 − 75.18 N
= 31.7 N (Ans.)
b) Find the total ver cal components of the forces.
ver cal component of ‘n’ of 60 N force = +60 N
ver cal component of 75 N force = +75 sin 62°
= +75 x 0.8829
= +66.22 N
ver cal component of 125 N force = −125 sin 55°
= −125 x 0.8192
= −102.39 N
ver cal component of 80 N force = −80 sin 20°
= −80 x 0.342
= −27.36 N
  

Total ver cal components = +60 + 66.22 − 102.39 − 27.36
= −3.53 N (Ans.)
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c) Find the resultant of these forces.
3.53
tan  = ____
 
  = 0.11356
31.7
 = tan−1 0.11356 = 6.35°
3.53
sin  = ____
      
R
3.53
3.53
or:
R = ________
 
    
   
= 31.9 N
= ______
 
sin 6.35° 0.1106

The resultant is: 31.9 N at 6.35° S of E (Ans.)
9. Two forces, ac ng on one point, are designated A and B. The components of Force A are XA = −300 N
and YA = 90 N. The components of Force B are XB = 220 N and YB = −60 N. Give a complete descrip on
of each force and find their equilibrant.
For Force A:
90
tan  = ____
     = 0.3
300
 = tan−1 0.3 = 16.7°
and


90
sin  =  ___  
A
90
90
    
A = ____
    = ______
 
= 313 N
sin  0.2873
Force A is: 313 N at 16.7° N of W (Ans.)
(ConƟnued)
• 3rd Class - Edition 2.0
9. (ConƟnued)
For Force B:
60
tan  = ____
     = 0.2727
220
 = tan−1 0.2727 = 15.3°
and


60
sin  =  ___  
B
60
60
    
B = ____
    =  ______
= 228 N
sin  0.2639
Force B is: 228 N at 15.3° S of E (Ans.)
To find resultant:
Total x components = 220 N − 300 N = −80 N
Total y components = 90 N − 60 N = +30 N
30
tan  =  ___  = 0.375
80
 = tan−1 0.375 = 20.56°
2
+ 802
R2 = 30
__________
R = √_____
900 + 6400  
R = √ 7300  = 85.44 N
 Resultant is 85.44 N at 20.56° N of W.
Since the equilibrant is directly opposing, it is described as 85.44 N at 20.56° S of E. (Ans.)
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10. A load is suspended by two cables, similar to the system shown in Example 7. However, cable A
makes an angle of 30° below the horizontal and cable B makes an angle of 50° below the horizontal.
If the tensions in A and B are 783 kN and 1055 kN respec vely, what is the downward force caused
by the load?
The load supported equals the sum of the ver cal components of A and B.
ver cal component of A
    
 
 
sin 30° =_____________________
 
783 kN

YA = 783 x sin 30° kN
= 783 x 0.5
= 391.5 kN
 ver cal component of  B
    
 
 
and
sin 50° = _____________________
 
1055 kN

YB = 1055 x sin 50° kN
= 1055 x 0.766
= 808.1 kN
 Force exerted by load = 391.5 + 808.1 kN
= 1199.6 kN (Ans.)
11. Find the force required to move a pump and the skid on which it rests. The total mass is 2000 kg. The
coefficient of sta c fric on between the floor and the skid is 0.26.
Normal force, RN = mass x 9.81 N
= 2000 x 9.81
= 19 620 N
Force required =  x RN
= 0.26 x 19 620 N
= 5101 N (Ans.)
12. Find the mass of an object when a horizontal force of 2100 N just starts to move the object along a
horizontal surface. The coefficient of sta c fric on is 0.40.



Force =  x RN
2100 N
Force ______
RN=  _____
 =  
 
   
  
0.4
= 5250 N
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but

 
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RN = mass x 9.81
R N
mass = _________
  N   
9.81 N/kg
5250 N
=  _________
   
9.81 N/kg
= 535 kg (Ans.)
13. Find the coefficient of fric on between the floor and a 1000 kg concrete pad if a force of 9000 N is
required to move the pad.
F
 =  ___   
RN
where

RN = mass x 9.81 N
9000 N
      
 = ____________
 
1000 x 9.81 N
9000
=  _____  
9810
= 0.92 (Ans.)
14. Find the mass of a body if a 200 N force moves it at contant speed with a fric on angle of 22°.
Given that F = 200 N; and fric on angle = 22°
F
then
tan 22° = ___
     
RN
F
200 N
or
RN = _______
 
    
  = 495 N
= _____
 
tan 22° 0.404
RN N
    
mass =  _________
9.81 N/kg
495
= ____
     kg
9.81
= 50.46 kg (Ans.)
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15. Find the mass of a plate if a force of 800 N, applied above the horizontal at the fric on angle, causes
mo on. The coefficient of fric on is 0.39.
First find the fric on angle, , using:
 = tan 

tan  = 0.39
 = tan−1 0.39 = 21.3°



Find the horizontal component of F:
Fx = F cos 21.3°
= 800 x 0.9317 N = 745.36 N
 
FF = Fx = 745.36 N
Find ver cal component of F:
Fy = F sin 21.3°
= 800 x 0.36 = 288 N
And this force li s up on the mass
  Actual normal force = force due to mass − 288 N
RN = force due to mass − 288 N
F 745.36
 
 F  = ______
 
but
RN = __
  
= 1911.18 N
0.39
 
1911.18 N = mass x 9.81 − 288 N
1911.18 + 288 N
= 224.2 kg (Ans.)
     
 
mass = ______________
 
9.81 N/kg
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16. Find the coefficient of fric on between two surfaces if the applied force of 950 N pulls at 28° above
the horizontal. The mass being moved by this applied force is 700 kg.
Horizontal component of F,
Fx = 950 x cos 28°
= 950 x 0.8829
= 838.76 N

FF = Fx = 838.76 N
Ver cal component of F,
Fy = 950 x sin 28°
= 950 x 0.4695
= 446.03 N
Downward force due to mass of 700 kg:
= m x 9.81
= 700 x 9.81 N = 6867 N

RN = Downward forces − Upward forces
= 6867 − 446.03 N
= 6421 N
F 838.76
 
  
= 0.13 (Ans.)
and
 =  ___F  = ______
RN 6421
17. A push of 600 N is applied to a block at a downward angle of 45° to the horizontal, which keeps the
block moving along the floor. If the coefficient of fric on between the block and the floor is 0.27,
what is the fric on angle?
tan (fric on angle)
=

tan  = 0.27
 = tan−1 0.27
= 15.1° (Ans.)
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18. If the coefficient of fric on between two surfaces is 0.21 and the mass of the pump to be pulled
along the floor is 800 kg, find the required pulling force when the force is directed at the following
upward angles from the horizontal: a) 5°, b) 11.9°, c) 17°.
a)
Downward force due to mass
= 800 x 9.81
= 7848 N
 = 0.21
Fx = F cos 5° = 0.9962 F
Fy = F sin 5° = 0.0871 F
since
forces left = forces right
then
FF = 0.9962 F
and
forces up = forces down
so
RN = 7848 N − 0.0871 F
since
 = 0.21
F
0.9962 F

0.21 = ___
  F  = ______________
      
 
RN 7848 − 0.0871 F
0.21(7848 − 0.0871 F) = 0.9962 F
1648.1 − 0.0183 F = 0.9962 F
1648.1 = 1.0145 F
1648.1
 
F =  ______ 
1.0145
= 1624.5 N (Ans.)
(ConƟnued)
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18. (ConƟnued)
b) 11.9°
Fx = F cos 11.9° = 0.9785 F
Fy = F sin 11.9° = 0.2062 F
forces le = forces right
so
FF = 0.9785 F
forces up = forces down
so
RN = 7848 − 0.2062 F
F
0.9785 F
      
 = 0.21 = ___
  F  = ______________
 
RN 7848 − 0.2062 F
(0.21)(7848 − 0.2062 F) = 0.9785 F
1648.1 − 0.0433 F = 0.9785 F
1648.1 = 1.0218 F
1648.1
 
  = 1613 N (Ans.)
F = ______
 
1.0218
c) 17°
Fx = F cos 17° = 0.9563 F
Fy = F sin 17° = 0.2924 F
so
FF = 0.9563 F
and
RN = 7848 − 0.2924 F
F
0.9563 F
      
 = ___
  F  = ______________
 
RN 7848 − 0.2924 F
(0.21)(7848 − 0.2924 F) = 0.9563 F
1648.1 − 0.0614 F = 0.9563 F
1648.1 = 1.0177 F
1648.1
 
  = 1619 N (Ans.)
F = ______
 
1.0177
19. Find the fric on angle in ques on 18.

tan of fric on angle = 
tan = 0.21
 
 = tan−1 0.21
= 11.9° (Ans.)
20. What angle of applied force will the least effort be required?
By defini on, the least effort is required at the fric on angle. (Ans.)
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