a. 10 units
b. 14 units
c. 16 units
d. 20 units
e. 24 units
Result: FR = (Fx2 + Fy2)1/2
= (62 + 82)1/2
= (36 + 64)1/2
= (100)1/2
= 10 N
2. The magnitude of vector A is 3 units and vector B is 4 units. If the addition of vectors A and B (A + B) is 5 units, the angle made by these two vectors is . . .
a. 30o
b. 45o
c. 60o
d. 73o
e. 90o
Result: R = (A2 + B2 + 2 A.B cos α)1/2
5 = (32 + 42 + 2.3.4 cos α)1/2
25 = (32 + 42 + 2.3.4 cos α)
25 = (9 + 16 + 24 cos α)
25 = 25 + 24 cos α
25 - 25 = 24 cos α
0 = 24 cos α
cos α = 0
α = arc cos-1 0
α = 90o
3. One vector of 10 units in magnitude directed to the right and another of 5 units in magnitude is directed to the left act on an object. To maintain the object at rest, we need an additional vector of . . .
a. 10 units in magnitude that is directed to the right
b. 15 units in magnitude that is directed to the left
c. 5 units in magnitude that is directed to the right
d. 5 units in magnitude that is directed to the left
e. zero length
Result: ∑F = 0
Fx+ + Fx- + Fr = 0 10 + (-5) + Fr = 0
Fr = 5 - 10
Fr = -5 (minus sign shows that the direction is to the left).
4. Look at the following scheme. The magnitude of the component of resultant vector to X and Y directions are . . .
b. Fx = (3√3 - 2√2), Fy = (3√2 + 2√2)
c. Fx = (3√3 + 2√2), Fy = (3 + 2√2)
d. Fx = (3√3 + 2√2), Fy = (3√2 + 2√2)
e. Fx = (3√3 - 2√2), Fy = (3√3 + 2√2)
Result: Fx = 6 units cos 30 + 4 units cos 135
Fx = 6 units 0.5√3 - 4 units 0.5√2
Fx = 3√3 - 2√2;
Fy = 6 units sin 30 + 4 units sin 135
Fy = 6 units 0.5 + 4 units 0.5√2
Fy = 3 + 2√2.
The answer is A.
5. Vector A of 20 units in magnitude makes an angle of 45o. with respect to the +X direction. The components of this vector in the X and Y directions are . . .
a. Ax = 20 units, Ay = 20 units
b. Ax = 20√2 units, Ay = 20√2 units
c. Ax = 10 units, Ay = 10 units
d. Ax = 10√2 units, Ay = 10√2 units
e. Ax = 10√3 units, Ay = 10√2 units
Result:
Ax = 20 units cos 45
Ax = 20 units 0.5√2
Ax = 10√2
Ay = 20 units sin 45
Ay = 20 units 0.5√2
Ay = 10√2
The answer is D.
6. Look at the following figure. The resultant force of the three vectors is . . .
b. 20 N
c. 30 N
d. 40 N
e. 70 N
Result:
Fx = 30 N cos 30 + 10 N cos 90 + 30 N cos 150
Fx = 30 N (0.5√3) + 10 N (0)
Fx = 15√3 - 15√3
Fx = 0;
Fy = 30 N sin 30 + 10 N sin 90 + 30 N sin 150
Fy = 30 N (0.5) + 10 N (1) + 30 N (0.5)
Fy = 15 + 10 + 15
Fy = 40
R = √(Fx2+Fy2)
= √(02+402)
= 40
The answer is D.
7. The resultant of the following three vectors is . . .
a. 3 units
b. 5 units
c. 5√5 units
d. 10 units
e. 10√3 units
Result:
Fx = 10units cos 60 + 5units cos 180 + 5units cos 300
Fx = 10units (0.5) + 5units (-1) + 5units (0.5)
Fx = 5 - 5 + 2.5
Fx = 2.5;
Fy = 10units sin 60 + 5units sin 180 + 5units sin 300
Fy = 10units (0.5√3) + 5units (0) + 5units (-0.5√3)
Fy = 5√3 + 0 - 2.5√3
Fy =2.5√3
R = √(Fx2+Fy2)
= √(2.52+2.5√32)
= 5
The answer is B.
8. Two vectors, F1 and F2, make an angle of 105o. The resultant of these two vectors makes an angle of 60o with respect to vector F2. If the magnitude of vector F2 is 8 units, the magnitude of vector F1 is . .
a. 10 units
b. 6 units
c. 4√6 units
d. 2√2 units
e.√6 units
Result:
F1 /sin α = F2 /sin β
F1 / sin (60) = 8 / sin (105 - 60)
F1 / sin (60) = 8 / sin (45)
F1 = 8 . sin (60) / sin (45)
F1 = 8 . 1/2√3 / 1/2√2
F1 = 4√6 units
The answer is C.
9. Look at the following figure. The resultant of the two vectors in the figure is . . .
a. 3 units
b. 6 units
c. 8 units
d. 10 units
e. 11 units
Result:
10. An object is placed on a smooth surface, It is pulled with a force of 20 N making an angle of 60 with respeect to the surface. If the mass of the object is negative, the norm of the vertical force needed to maintain the object at rest is . . . N
a. 20√3
b. 20√2
c. 20
d. 10√3
e. 10√2
Result:
11. Three forces A, B, and C, act on a massless body simultaneously, as shown in the following figure. The component of resultant force in the X and Y directions is . . .
a. Rx = 5√3 units, Ry = 5√2 units
b. Rx = 0 units, Ry = 5 units
c. Rx = 5√3 units, Ry = -5√2 units
d. Rx = 5 units, Ry = 10 units
e. Rx = -5√3 units, Ry = 15 units
Result:
Fx = A cos 30 + B cos 210 + C cos 150
Fx = 10 cos 30 + 5 cos 210 + 5 cos 150
Fx = 10 (0.5√3) + 5 (-0.5√3) + 5 (-0.5√3)
Fx = 5√3 - 2.5√3 - 2.5√3
Fx = 0;
Fy = A sin 30 + B sin 210 + C sin 150
Fy = 10 (0.5) + 5 (-0.5) + 5 (0.5)
Fy = 5 - 2.5 + 2.5
Fy =5
R = √(Fx2+Fy2)
= √(02+52)
= 5
The answer is B.
12. Vector A has 6 units in magnitude and vector C has 5 units in magnitude. If A + B = 2C and the angle made by vectors A and B is 90o. The magnitude of vector B is . . .
a. 4 units
b. 6 units
c. 8 units
d. 10 units
e. 25 units
Result:
13. Vector B is 10 N in magnitude and vector C is 20 N in length. If A + B = C and the angle between A and C is 30o. The angle between vectors B and C is . . .
a. 30o
b. 45o
c. 60o
d. 90o
e. 120o
Result:
14. An object is pulled to the right with a force of 30 N that makes an angle of 30o with respect to the surface of the floor. The magnitude and direction of an additional vector that must be given to the object to maintain the object at rest is . . .
a. 20 N to the right
b. 24 N to the left
c. 10 N to the right
d. 10 N to the left
e. 18 N to the up
Result:
15. Two vectors have the same magnitude. If the fraction of addition and subtraction of these vectors is √3, the angle made by these two vectors is . . .
a. 30o
b. 37o
c. 45o
d. 60o
e. 120o
Result:
II. STRUCTURED QUESTIONS
Answer all the questions on the answer sheet provided. Calculators are allowed.
1. The figure below shows two vectors, A and B. Determine the resultant.
4. Vector B has a magnitude of 20 N and vector C has 25 N in length. If A + B = C and the angle between vectors A and C is 30°, determine the angle between vectors B and C.
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