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The concept of friction, static friction, kinetic friction, and centripetal force, with examples and problems. It also covers the forces acting on objects in circular motion and the source of centripetal force. calculations for the force required to move a box, the accelerations of blocks on an incline, and the motion of a ball on a string.
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6. Forces and Motion-II
Friction:
F f
F f
Static Friction: N
f mg
F
s
∑ Fy =^ N^ −^ mg^ −^ Fsin^ θ^ =^0 N (^) = mg (^) + F sin θ ∑ Fx =^ F^ cos θ^ −^ f^ s >^0 F cos θ (^) − μ s N (^) > 0
F cos θ (^) − μ s Fsin θ (^) > μ smg
F (^) > μ smg cos θ (^) − μ s sin θ
Kinetic Friction: Once an object is in motion, the friction force changes to kinetic friction force, which is proportional to the magnitude of normal force, f (^) k = μ k N
where μ k is the coefficient of kinetic friction.
is easier to keep it moving.
Problem 39 (p. 127): A 10 kg block rests on top of a 40 kg slab which rests on a frictionless floor. The coefficient of static and kinetic friction between the block and the slab are 0.60 and 0.40 respectively. The block is pulled with a force of 100 N. What are the accelerations of (a) the block and (b) the slab?
(^100) m
m
y
x
1 2 N F (^) f
m 1 g
N f
m 2 g
N
'
N
a 1 = μ k g −
m 1 = −6.1 m / s^2 Slab: (^) ∑Fx = − μ k N (^) = m 2 a 2 − μ k m 1 g^ = m 2 a 2
a 2 = − μ k m 1 g m 2 = − 0. 98 m / s^2
Example: Two identical blocks of 10 kg each are sitting on an incline with an angle θ = 30°. One block is released so that its initial speed is zero, while the other block is released with an initial speed of 1 m/s. What are the accelerations of each block just after the release and 10 s later? The coefficients of static and kinetic friction are 0.7 and 0.6. N (^) f
θ mg
θ
0 =^ N^ f
θ mg
θ
= 1 y
x
v 0 v (^0) x m/s
Block 1 : (^) ∑ Fy = N (^) − mg cos θ (^) = 0
N (^) = mg cos θ f (^) s = μ s N = μ smg^ cos θ = 59 N Fgx = mgsin θ = 49 N < f (^) s The block will remain stationary.
Block 2 : (^) ∑ Fy = N (^) − mg cos θ = 0
N (^) = mg cos θ ∑ Fx =^ mgsin^ θ^ −^ f^ =^ ma mgsin (^) θ − μ k N (^) = ma mgsin (^) θ − μ k mg cos θ = ma a = g(sin θ − μ k cos θ ) = −0.19 m / s^2
d (^) m
v k k
0 1 1
2 2
= 5
μ μ
θ θ
N (^) f
θ mg
θ
y
x 1 1
k
∑ Fy = N − mg cos θ 1 = 0
N = mg cos θ 1
∑ Fx = mgsin θ 1 − fk = max 1
mg sin θ 1 − μ k mg cos θ 1 = max 1
ax 1 =^ g(sin θ 1 −^ μ k cos θ 1 )
= (9.8 m / s^2 )(sin 30°−0.6 cos30°) = −0.19 m / s^2 x = x 0 + v (^0) x t 1 + 12 ax 1 t 12
2 =^0 +^1 ×^ t 1 −^12 (0.19)t 12
0.092t 12 − t 1 + 2 = 0
t 1 = 1.6 or −12.1 s
v =^ v 0 +^ ax 1 t 1
= 1 − 0.19 ×1. = 0.70 m / s
∑ Fy = N^ − mg^ cos θ 2 = 0 N (^) = mg cos θ 2 ∑ Fx = − f^ k − mgsin^ θ 2 = max 2 − μ k 2 mg^ cos θ 2 − mgsin^ θ 2 = max 2 μ k 2 gcos θ 2 = −gsin θ 2 − ax 2
μ k 2 =
−gsin^ θ 2 − ax 2 g cos θ 2 = 0.^15
H We solve this complicated problem one piece at a time. v^2 =^ v 02 +^2 a(x −^ x 0 ) 0 =^ v 02 +^2 ax 2 (d −^ 0)
d = −^ v 02 2 ax 2
2 2(−2.33) = (^) 0.11 m
Centripetal Force: The force directed toward the center that provides the acceleration for an object in a circular motion:
Fc = ma = m v^2 r e.g. gravity, tension, normal force.
Example: A 2 kg ball on a string of length L = 1 m has a period of 0.5 s. Ignoring the gravitational force, what is the speed, centripetal acceleration, and tension?
Problem 5 7 (p.128) A mass m on a frictionless table is attached to a hanging mass M by a cord through a hole in the table. Find the speed with which m must move in order for M to stay at rest for a radius r.
M
m
r T
Mg
T
T = mac = m v^2 r T − Mg = 0
Mg =^ m v^2 r
v =^ Mgr m
Conceptual Question:
Determine the source of the centripetal force that causes the object to move in a circle:
r
Conceptual Question:
(a) v (^) = 2 π r T
=
2 π r π 3 = 6 r = 0.3 m s
a (^) = v^2 r = 1.82 m s^2 (b) f (^) = ma
= 3.^64 × 10 −^3 N
(c) At r = 10 cm, the friction is just enough to provide the centripetal force:
μ N (^) = mv^2 r
μ mg (^) = mv
2 r
μ (^) = v^2 gr
2 π r T
2
gr
= 4 π^2 r gT 2
=
4 π^2 r
g π 3
2
36 r g = 0.^37
