3.49
The 0.1-kg particle has a speed v = 10 m/s as it pass the 30◦ position shown. The coefficient of kinetic
friction between the particle and the vertical plane track is µk = 0.20. Determine the magnitude of the total
force exerted by the track on the particle. What is the deceleration of the particle?
Solution:
Summing the forces in the normal direction,
0 = mg cos θ + N − man
N = man − mg cos θ
2
v
N =m
− mg cos θ
ρ
2 !
10 ms
N = 0.1 kg
− 0.1 kg cos (30◦ )
5m
(1)
(2)
(3)
(4)
N = 1.15 N
(5)
Total force exerted by track is normal force plus friction force.
q
q
2
Ftotal = N 2 + (µk N ) = N 1 + µ2k = 1.17 N
(6)
Summing the forces in the tangential direction,
0 = −mg sin θ − µk N − mat
N
at = −g sin θ − µk
m
m
at = −7.21 2
s
(7)
(8)
(9)
3.89
The 3000-lb car is traveling 60 mi/hr on the straight portion of the road, and then its speed is reduced
uniformly from A to C, at which point it comes to rest. Compute the magnitude F of the total friction force
exerted by the road on the car
a. just before it passes point B.
b. just after it passes point B.
c. just before it stops at point C.
Solution:
For constant acceleration,
v 2 − vo2 = 2a (s − so )
2
vC
2
vA
−
2 (sC − sA )
sC − sA = 100 ft + 30/180 ∗ π ∗ 250 ft = 230.9 ft
mi 1 hr 5280 ft
ft
vA = 60
= 88.0
hr 3600 s 1 mi
s
vC = 0
a=
(88 fts )2
(0)2 −
2(230.9 ft)
ft
a = −16.7 2
s
2
2
vB
= vA
+ 2a (sB − sA )
ft
ft
vB = sqrt(88 )2 − 2(16.7 2 ) (100 ft)
s
s
ft
vB = 66.3 2
s
a=
(10)
(11)
(12)
(13)
(14)
(15)
(16)
(17)
(18)
(19)
To find the force, the total acceleration is needed. F = ma. Before B all there is is tangential acceleration.
Just before B the total friction force is:
3000
ft
FF,a = ma =
slug 16.7 2 = 1.56 × 103 lb
(20)
32.2
s
Just after B there is a normal component of the acceleration.
anb = VB2 /ρ =
(66.3 fts )2
ft
= 17.6 2
250 ft
s
The total acceleration just after B is
r
atotal,b =
FF,b
ft 2
ft
ft
) + (17.6 2 )2 = 24.3 2
2
s
s
s
= ma = 2.26 × 103 lb
(16.7
(21)
(22)
(23)
At C there is no normal component of the acceleration because the speed is zero.
FF,c = FF,a = 1.56 × 103 lb
(24)