Physics 151 Class Exercise: Review of Forces
1.
A picture hangs on the wall suspended by two strings as shown below. The tension in string
1 is 1.7 N.
(a) Is the tension in string 2 greater than, less than, or equal to 1.7 N? Explain.
(b) Verify your answer to part (a) by calculating the tension in string 2.
(c) What is the weight of the picture?
T1
T2
y
65º
x 32º
W
Free Body Diagram w/Coordinate System
(a)
(b)
(c)
Less; horizontal components of the two string tensions must be equal, and cosθ 2 > cosθ1.
ΣFx = T2 cos θ 2 − T1 cos θ1 = 0
T1 cos θ1 = T2 cos θ 2
cos θ1
T2 = T1
cos θ 2
cos 65°
= (1.7 N)
cos 32°
= 0.85 N
ΣFy = T1 sin θ1 + T2 sin θ 2 − W = 0
W = T1 y + T2 y = T1 sin θ1 + T2 sin θ 2 = (1.7 N)sin 65° + (0.85 N)sin 32° = 2.0 N
2.
A popular ride at amusement parks is illustrated below. In this ride, people sit in a swing that
is suspended from a long rotating arm. Riders are at a distance of 12 m from the axis of rotation
and move with a speed of 25 mi/h.
(a) Find the centripetal acceleration of the riders.
(b) Find the angle θ of the supporting wires make with the vertical.
(c) Notice that the swings shown are at the same angle to the vertical regardless of the weight of the
rider. Explain.
T
θy
x
W
Free Body Diagram w/Coordinate System
(a) acp =
2
v
=
r
⎡
⎣⎢
( )(
25 mi
h
1609 m
mi
)(
1h
3600 s
)
12 m
⎤
⎦⎥
2
= 10 m/s 2
(b) Taking T as the chain tension,
= T sin θ = macp
= T cos θ − mg = 0
= macp
= mg
acp
tan θ =
g
⎛ acp ⎞
θ = tan −1 ⎜⎜
⎟⎟
⎝ g ⎠
⎛ 10.4 m2 ⎞
−1 ⎜
s ⎟
= tan
⎜ 9.81 m2 ⎟
s ⎠
⎝
= 47°
ΣFx
ΣFy
T sin θ
T cos θ
Dividing one equation by the other
(c) The angle corresponds to a ratio of vertical to horizontal force. Both of those forces are
proportional to m, so that m drops out of the ratio.