1. More Vector Examples Answers
1. A box is being pulled across the floor by a rope at an angle of 30o to
the horizontal. The tension in the rope is 50 newtons.
(a) Determine Fh , the horizontal component of the force.
adj
50 N
cos 30 =
30o
hyp Fh Fh
50 N
SOH-CAH-TOA
(50 N) =
50 N
Fh = 43 N

2. More Vector Examples Answers
1. A box is being pulled across the floor by a rope at an angle of 30o to
the horizontal. The tension in the rope is 50 newtons.
(b) Determine Fv, the vertical component of the force on the rope.
opp
50 N Fv
sin 30 =
30o
hyp
43 N Fv
50 N
(50 N) =
SOH-CAH-TOA
50 N
Fv = 25 N

3. 2. A plane is flying with a speed of 150 m/s at a bearing of 120o.

4. 2. A plane is flying with a speed of 150 m/s at a bearing of 120o.ve
30o
180o
150 m/s
(a) Determine the eastward component of the velocity.
adj
cos 30 =
hyp ve
150 m/s
ve = 130 m/s
(150 m/s) =
150 m/s

5. 2. A plane is flying with a speed of 150 m/s at a bearing of 120o.vs
180o
150 m/s
(b) Determine the southward component of the velocity.
opp
sin 30 =
hyp vs
150 m/s
vs = 75 m/s
(150 m/s) =
150 m/s

6. 2. A plane is flying with a speed of 150 m/s at a bearing of 120o.
(c) How far east has the plane flown after 10 seconds?
d = v t
de = ve t
= ( 130 m/s )( 10 s )
de = m

7. 3. A rope is tied to a box on the floor and is held at angle of 40o to
the horizontal. The box has a mass of 5.0 kg. A force of
60 newtons is exerted on the box via the rope.
(a) Determine the horizontal and vertical components of the force.
60 N
40o
5.0 kg

8. 3. A rope is tied to a box on the floor and is held at angle of 40o to
the horizontal. The box has a mass of 5.0 kg. A force of
60 newtons is exerted on the box via the rope.
(a) Determine the horizontal and vertical components of the force.
60 N Fv
40o
5.0 kg Fh
Fh :
FV :
opp
adj
sin 40 =
cos 40 =
hyp
hyp Fh
60 N Fv
60 N
(60 N) =
(60 N) =
60 N
60 N
Fh = N
Fv = N

9. 3. (b) Calculate the normal force.
5.0 kg
46.0 N
FN = ?
Wt.
= m g
= ( 5.0 kg )( 9.8 m/s2 ) = 49 N

10. 3. (b) Calculate the normal force.
5.0 kg
46.0 N
FN = ?
Wt. = 49 N
Fup = Fdown
FN + Fv = Wt.
FN N = 49 N N N
FN = N

11. 3. (c) If the coefficient of friction is 0
3. (c) If the coefficient of friction is 0.60, find the acceleration of
the box.
60 N
40o Ff
5.0 kg
46.0 N
a = ?
μ = 0.60
10.4 N

12. 3. (c) If the coefficient of friction is 0
3. (c) If the coefficient of friction is 0.60, find the acceleration of
the box.
60 N
40o Ff
5.0 kg
46.0 N
a = ?
Fnet = ?
μ = 0.60
10.4 N
Fnet
Fnet = m a
a = m m m
Fnet = Fh + Ff
Ff = μ FN
= ( 0.60 )( 10.4 N )
= N

13. 3. (c) If the coefficient of friction is 0
3. (c) If the coefficient of friction is 0.60, find the acceleration of
the box.
60 N
40o
5.0 kg
46.0 N
a = ?
6.3 N
Fnet = ?
μ = 0.60
10.4 N
Fnet
39.76 N
Fnet = m a
a = = m
5.0 kg
a = 8.0 m/s2
Fnet = Fh + Ff
= ( N ) + ( N )
= N