Application of Newton’s First Law of Motion (Examples)
(Problems taken from Young & Freedman, University Physics, 13 th ed.)
A. Object at Rest
Example 6. A large wrecking ball is held in place by two light steel
cables as shown in Fig. 2. If the mass m of the ball is 4090 kg, what
are the (a) tension TB in the cable that makes an angle of 40 o with
the vertical and (b) tension TA in the horizontal cable?
Solution:
Figure 1.3. Example 6
A. Set-up: Draw FBD for m.
TBy
TB
40o
TA
TBx
●
Fg
B. Execute: for horizontal forces
or vertical forces
∑ 𝐹𝑥 = 0
∑ 𝐹𝑦 = 0
𝑇𝐵𝑥 − 𝑇𝐴 = 0
𝑇𝐵𝑦 − 𝐹𝑔 = 0
where 𝐹𝑔 = 𝑚𝑔 and 𝑇𝐵𝑦 = 𝑇𝐵 𝑐𝑜𝑠40𝑜
where 𝑇𝐵𝑥 = 𝑇𝐵 𝑠𝑖𝑛40𝑜
𝑇𝐵 𝑠𝑖𝑛40𝑜 − 𝑇𝐴 = 0
𝑇𝐵 𝑠𝑖𝑛40𝑜 = 𝑇𝐴
Equate (1) and (2):
𝑇𝐵 𝑐𝑜𝑠40𝑜 − 𝑚𝑔 = 0
𝑇𝐵 𝑐𝑜𝑠40𝑜 = 𝑚𝑔
(1)
𝑇𝐵 𝑠𝑖𝑛40𝑜
𝑇𝐴
=
𝑇𝐵 𝑐𝑜𝑠40𝑜 𝑚𝑔
𝑇𝐴 = 𝑚𝑔𝑡𝑎𝑛40𝑜 = (4090𝑘𝑔) (9.8 𝑘𝑔
From (2)
𝑚
) (𝑡𝑎𝑛40𝑜 ) = 𝟑. 𝟑𝟔𝒙𝟏𝟎𝟒 𝑵
𝑠2
𝑚
(4090𝑘𝑔) (9.8 𝑘𝑔 2 )
𝑚𝑔
𝑠
𝑇𝐵 =
=
= 𝟓. 𝟐𝟑𝒙𝟏𝟎𝟒 𝑵
𝑐𝑜𝑠40𝑜
𝑐𝑜𝑠40𝑜
(2)
Example 7. A 1130-kg car is held in place by a light cable on a
very smooth frictionless ramp, as shown in Fig. 3. The cable
makes an angle of 31.0o above the surface of the ramp, and the
ramp itself rises at 25 o above the horizontal. (a) Draw FBD for
the car. (b) Find the tension in the cable. (c) How hard does the
surface of the ramp push on the car?
Figure 1.4. Example 7
car held in place
Solution:
(a) Set-up: Draw FBD for the car.
T – tension in cable
FN – normal force
T
FN
Tsin31o
●
Tcos31o
mgsin25o
mgcos25o
mg
Execute:
(b) Net force on car
(parallel to the inclined plane)
∑ 𝐹𝑝𝑎𝑟𝑎𝑙𝑙𝑒𝑙 = 0
𝑇𝑐𝑜𝑠 31𝑜 − 𝑚𝑔𝑠𝑖𝑛25𝑜 = 0
𝑚𝑔𝑠𝑖𝑛25𝑜
𝑐𝑜𝑠 31𝑜
𝑚
(1130𝑘𝑔) (9.8𝑘𝑔 2 ) 𝑠𝑖𝑛25𝑜
𝑠
𝑇=
𝑐𝑜𝑠31𝑜
𝑇=
𝑻 = 𝟓𝟒𝟔𝟎𝑵
(c) Net force on car
(perpindicular to the inclined plane)
∑ 𝐹𝑝𝑒𝑟𝑝𝑖𝑛𝑑𝑖𝑐𝑢𝑙𝑎𝑟 = 0
𝐹𝑁 + 𝑇𝑠𝑖𝑛 31𝑜 − 𝑚𝑔𝑐𝑜𝑠25𝑜 = 0
𝐹𝑁 = 𝑚𝑔𝑐𝑜𝑠25𝑜 − 𝑇𝑠𝑖𝑛 31𝑜 = 0
𝑭𝑵 = 𝟕𝟐𝟐𝟎𝑵
FN refers to the force the surface of the ramp
exerts on the car.
B. Objects Moving with Constant Velocity
Example 8. A block with mass m 1 is placed on an inclined
plane with slope angle 𝛼 and is connected to a second
hanging block with mass m 2 by a cord passing over a small,
frictionless pulley. The coefficient of static friction is 𝜇𝑠 and the
coefficient of kinetic friction between m 1 and inclined plane is
𝜇𝑘 .
(a) Find the mass m2 for which m1 moves up the plane at
Figure 5. Example 8
constant speed once it is set in motion.
(b) Find the mass m2 for which m1 moves down the plane at constant speed once it is set in
motion.
Solution:
Set-up: Draw FBD for each block.
a) m1 moves up the plane at constant
speed
FN
T
T
fk
●
mgsin𝛼
mgcos25o
●
m1g
m2g
m1g
Block m1
Block m2
where T – tension in cord
FN – normal force
𝐹𝑁 = 𝑚1 𝑔𝑐𝑜𝑠25𝑜
𝑓𝑘 = 𝜇𝑘 𝐹𝑁 = 𝑚1 𝑔𝑐𝑜𝑠25𝑜
Execute:
Execute:
Net force on m2 (parallel to the inclined plane)
Net force on m2
∑ 𝐹𝑝𝑎𝑟𝑎𝑙𝑙𝑒𝑙 = 0
∑ 𝐹𝑝𝑎𝑟𝑎𝑙𝑙𝑒𝑙 = 0
𝑇 − 𝑚1 𝑔𝑠𝑖𝑛𝛼 − 𝑓𝑘 = 0
𝑇 − 𝑚2 𝑔 = 0
𝑇 − 𝑚1 𝑔𝑠𝑖𝑛𝛼 − 𝜇𝑘 𝑚1 𝑔𝑐𝑜𝑠25𝑜 = 0
𝑇 = 𝑚1 𝑔𝑠𝑖𝑛𝛼 + 𝜇𝑘 𝑚1 𝑔𝑐𝑜𝑠25𝑜 = 0
Equate (1) and (2):
(1)
𝑇 = 𝑚2 𝑔
(2)
𝑚2 𝑔 = 𝑚1 𝑔𝑠𝑖𝑛𝛼 + 𝜇𝑘 𝑚1 𝑔𝑐𝑜𝑠25𝑜
𝑚2 = 𝑚1 𝑠𝑖𝑛𝛼 + 𝜇𝑘 𝑚1 𝑐𝑜𝑠25𝑜 = 𝑚1 (𝑠𝑖𝑛𝛼 + 𝜇𝑘 𝑐𝑜𝑠25𝑜 )
(b)
m1 moves down the plane at constant speed once it is set in motion.
FBD for m1:
FN
T
fk
mgcos25o
mgsin𝛼
m1g
m1g
∑ 𝐹𝑝𝑎𝑟𝑎𝑙𝑙𝑒𝑙 = 0
𝑇 − 𝑚1 𝑔𝑠𝑖𝑛𝛼 + 𝑓𝑘 = 0
𝑇 − 𝑚1 𝑔𝑠𝑖𝑛𝛼 + 𝜇𝑘 𝑚1 𝑔𝑐𝑜𝑠25𝑜 = 0
𝑇 = 𝑚1 𝑔𝑠𝑖𝑛𝛼 − 𝜇𝑘 𝑚1 𝑔𝑐𝑜𝑠25𝑜 = 0
(3)
Equate (2) and (3)
𝑚2 𝑔 = 𝑚1 𝑔𝑠𝑖𝑛𝛼 − 𝜇𝑘 𝑚1 𝑔𝑐𝑜𝑠25𝑜
𝑚2 = 𝑚1 𝑠𝑖𝑛𝛼 − 𝜇𝑘 𝑚1 𝑔𝑜𝑠25𝑜 = 𝑚1 (𝑠𝑖𝑛𝛼 − 𝜇𝑘 𝑐𝑜𝑠25𝑜 )