Newton's Laws and Frictional Forces
Answer Section
PROBLEM
1.ANS:
(a) Trigonometric Method
Looking at the vector triangle: = 20.
Using cosine law: = 1.3 N
Using sine law: = 18.
As a result: .
(b) Component Method
F1X = 3.5 N(cos 30) [E] = 3.03 N [E]
F2X = 2.8 N(sin 40) [W] = 1.80 N [W]
FX = 3.03 N [E] + 1.80 N [W] = 1.23 N [E]
F1Y = 3.5 N(sin 30) [N] = 1.75 N [N]
F2Y = 2.8 N(cos 40) [S] = 2.14 N [S]
FY = 1.75 N [N] + 2.14 N [S] = 0.39 N [S]
Using Pythagoras:
Using a trigonometric ratio: = tan–1 = 18.
As a result: .
The two methods give equivalent results.
REF:COBJ:2.3LOC:FM1.01
2.ANS:
(a)
Free-body diagram: / FN acting upFg acting down
FA acting as illustrated
FK acting to the right
“Up” and “to the right” are the positive directions.
Horizontally:
The acceleration of the object is 1.0 m/s2.
(b)
Vertically:
The normal force is 1.3 102 N[up].
(c)
Free-body diagram: / FN acting upFg acting down
FA acting to the left
FK acting to the right
“Up” and “to the right” are the positive directions.
The acceleration of the two masses is 0.59 m/s2.
REF:K/UOBJ:2.3LOC:FM1.02
3.ANS:
Using the component method of finding the total force exerted by strand 1, 2, and 3:North–South components: / Strand 1: 21 mN(cos 20) = 19.7 mN [N]
Strand 2: 16 mN(cos 60) = 8.0 mN [S]
Strand 3: 18 mN(cos 40) = 13.8 mN [S]
Total: 2.1 mN [S]
East–West components: / Strand 1: 21 mN(sin 20) = 7.2 mN [E]
Strand 2: 16 mN(sin 60) = 13.9 mN [E]
Strand 3: 18 mN(sin 40) = 11.6 mN [W]
Total: 9.5 mN [E]
Using Pythagoras and simple trig. ratio; the total force acting by the three strands is
9.7 mN [78ºE of S].
Since the spider web is in static equilibrium, the force exerted by strand 4 must be
9.7 mN [78ºW of N].
REF:K/UOBJ:2.3LOC:FM1.01
4.ANS:
(a)
Consider the system of the packages:
Free-body diagram: / FTsin 75 acting upFg acting down
FK acting down
At a constant speed, the acceleration and net force are both zero.
Using the sign convention that “up” is (–) and “down” is (+):
The force the winch must exert is 5.7102 N [up].
(b)
Consider the 10-kg package:
Free-body diagram: / FT acting upFg acting down
The force of tension in the cable connecting the two packages is 88 N [up].
(c)
Consider the 10-kg package:
Free-body diagram: / FT acting upFg acting down
The force of tension in the cable connecting the two packages is 108 N [up].
REF:K/UOBJ:2.3LOC:FM1.01
5.ANS:
(a)
For the 5.0-kg mass:
Free-body diagram: / FN acting perpendicular to ramp and upFg acting down
FT acting up along the ramp (this is the positive direction)
FK acting down along the ramp (this is the negative direction)
5.0 kg(a) = FT – mg(cos ) – mg(sin )
5.0 kg(a) = FT – 35.5 N
For the 20.0-kg mass:
Free-body diagram: / FT acting up (this is the negative direction)Fg acting down (this is the positive direction)
20.0 kg(a) – 196 N – FT
Solving the system of equations:
a = 6.4 m/s2
The acceleration of the 5.0-kg mass along the ramp is 6.4 m/s2.
(b)
The tension in the cable is 68 N.
(c)
The speed of projection of the mass off the top of the ramp is 7.2 m/s.
(d)
Vertically: Let “up” be (–) and “down” be (+).
a = 9.8 m/s2
d = 6.0 m
Horizontal range:
The horizontal range for the projected mass is 9.5 m.
REF:K/UOBJ:2.3LOC:FM1.01
6.ANS:
(a)
When the force is acting horizontally:
Free-body diagram: / FN acting upFg acting down
FA acting to the right
FK acting to the left
When the force is acting elevated:
Free-body diagram: / FN acting up
Fg acting down
FA acting 30 above the horizontal
FK acting to the left
(b)
Let “to the right” and “up” be (+).
Consider box when applied force is acting horizontally:
= 5.0 kg(1.2 m/s2) – 12 N
= –6.0 N
= mg
= 5.0 kg(9.8 N/kg)
= 49 N
The coefficient of sliding (kinetic) friction is 0.12.
(c)
When the force is acting 30º above the horizontal:
Vertically:
= –(–12 N(sin300) + 5.0 kg(9.8 N/kg))
= –43 N
Horizontally:
FK = FN
= 0.12(43 N)
FK = 5.16 N
= 12 N(cos30º) – 5.16 N
= 5.23 N
The acceleration of the box is 1.0 m/s2.
REF:K/UOBJ:2.3LOC:FM1.01
7.ANS:
(a)
Free-body diagram: / FN acting perpendicular to the roof (upward)Fg acting down
FK acting up along the roof (this is the negative direction)
(b)
Parallel to the roof:
ma = mg(sin) – mg(cos )
a = 9.8 N/kg(sin 50º) – (0.14)(9.8 N/kg)(cos 50º)
a = 6.62 m/s2
The ice leaves the roof at 1.0 101 m/s.
(c)
When the ice leaves the roof it becomes a projectile:
Vertically:
Solving the quadratic: t = 0.406 s
Horizontally:
d = v(cos t
= 10.3 m/s(cos 50)(0.406 s)
d = 2.7 m
The ice lands 2.7 m from the base of the building.
REF:K/UOBJ:2.3LOC:FM1.01
8.ANS:
(a)
For the 0.80-kg mass:Free-body diagram: / FN acting up
Fg acting down
FT acting to the right (this is the positive direction)
FK acting to the left (this is the negative direction)
0.80 kg(a) = FT – KFN
0.80 kg(a) = FT – 0.14(0.80 kg)(9.8 N/kg)
0.80 kg(a) = FT – 1.10 N
For the 2.0-kg mass:Free-body diagram: / FN acting perpendicular to the ramp (upward)
Fg acting down
FT acting up along the ramp (this is the negative direction)
FK acting up along the ramp
2.0 kg(a) = 2.0 kg(9.8 N/kg)(sin 30º) – FT – 0.14(2.0 kg)(9.8 N/kg)(cos 30º)
2.0 kg(a) = –FT + 7.42 N
Solving the system of equations: a = 2.3 m/s2
The system will accelerate at 2.3 m/s2.
(b)
FT = 0.80 kg(a) + 1.10 N
= 0.80 kg(2.26 m/s2) + 1.10 N
FT = 2.9 N
The tension in the string is 2.9 N.
(c)
If the block remains stationary:
FS = Fg sin
= 2.0 kg(9.8 N/kg)(sin 30)
FS = 9.8 N
The minimum coefficient of static friction required is 0.58.
REF:K/UOBJ:2.3LOC:FM1.01
9.ANS:
(a)
For the 4.0-kg mass:
Free-body diagram: / FN acting perpendicular to the ramp (upward)Fg acting down
FT acting up along the ramp (this is the positive direction)
FK acting down along the ramp (this is the negative direction)
For the 6.0-kg mass:
Free-body diagram: / Fg acting down (this is the positive direction)FT acting up (this is the negative direction)
(b)
For the 4.0-kg mass:
4.0 kg(a) = FT – mg(cos ) – mg(sin)
4.0 kg(a) = FT – 13.5 N
For the 6.0-kg mass:
6.0 kg(a) = 58.8 N – FT
Solving the system of equations:
a = 4.5 m/s2
The acceleration of the 4.0-kg mass along the ramp is 4.5 m/s2.
(c)
FT = 4.0 kg(a) +13.5 N
= 4.0 kg(4.53 m/s2) + 13.5 N
FT = 32 N
The tension in the cable is 32 N.
(d)
For the block sliding down the ramp:
Free-body diagram: / FN acting perpendicular to the ramp (upward)Fg acting down
FK acting up along the ramp (this is the negative direction)
ma = mg(sin ) – mg(cos )
a = 9.8 N/kg(sin 30º) – (0.18)(9.8 N/kg)(cos30º)
a = 3.37 m/s2
It would take 1.3 s to reach the bottom of the ramp.
REF:K/UOBJ:2.3LOC:FM1.01