HW4 Chapter 4
#1. Chapter 4: #42: A 2.00‐kg block is held in equilibrium on r
incline of angle θ = 60.0° by a horizontal force F applied in the an direction shown in Figure P4.42. If the coefficient of static friction between block and incline is μs = 0.300, determine (a) the minimum value of and (b) the normal force exerted by the incline on the block. 4.42
r
When the minimum force F is used, the block tends to
r
slide down the incline so the friction force, f s is directed
+y
up the incline.
Force
ur
fs
ur
n
F(cos60º)
+x
60.0°
ur
F
m
F(sin60º)
60.0°
Friction on box
by surface
+fx=μn
0
Normal on box
by surface
0
n
Weight on leg
by earth
Wx=-16.97N
mg = 19.6 N
using small n
so looks diff
than newton
Wy=-9.8N
While the block is in equilibrium, we have
ΣFx = F cos60.0° + f s − (19.6 N )sin 60.0° = 0
(1)
ΣFy = n − F sin 60.0° − (19.6 N )cos 60.0° = 0
(2)
For minimum F (impending motion), f s = ( f s )max = μ s n = ( 0.300 )n
(3)
Equation (2) gives n = 0.866 F + 9.80 N
(4)
(a) Equation (3) becomes: f s = 0.260 F + 2.94 N , so Equation (1) gives
0.500 F + 0.260 F + 2.94 N − 17.0 N = 0 , or F = 18.5 N
(b) Finally, Equation (4) gives the normal force
n = 25.8 N
24. An object with mass m1 = 5.00 kg rests on a frictionless horizontal and is connected to a cable that passes over a pulley and is then fastened hanging object with mass m2 = 10.0 kg, as shown in Figure P4.24. Find Figure P4.24 acceleration of each object and the tension in the cable. table to a the 4.24
First consider the block moving along the
horizontal. The only force in the direction
of movement is T. Thus,
ΣFx = ma x ⇒ T = ( 5.00 kg ) a
5.00 kg
ur
T
+y
(1)
Next consider the block which moves
vertically. The forces on it are the tension
T and its weight, 98.0 N.
ΣFy = ma y ⇒ 98.0 N − T = (10.0 kg ) a
ur
T
+x
10.0 kg
mg = 98.0 N
(a)
(b)
(2)
Note that both blocks must have the same magnitude of acceleration. Equations (1) and
(2) can be solved simultaneously to give.
a = 6.53 m s2 , and T = 32.7 N
#3. Chapter 4: 32: A hockey puck is hit on a frozen lake and starts moving with a speed of 12.0 m/s. Five seconds later, its speed is 6.00 m/s. (a) What is its average acceleration? (b) What is the average value of the coefficient of kinetic friction between puck and ice? (c) How far does the puck travel during the 5.00‐s interval? 4.32
(a)
ax =
v x − v0 x 6.00 m s − 12.0 m s
=
= − 1.20 m s2
t
5.00 s
(b) From Newton’s second law, ΣFx = − f k = ma x , or f k = − ma x .
The normal force exerted on the puck by the ice is n = mg , so the coefficient of
friction is
2
f k − m ( −1.20 m s )
μk = =
= 0.122
n
m ( 9.80 m s2 )
(c)
⎛ v + v ⎞ ⎛ 6.00 m s + 12.0 m s ⎞
Δx = ( v x )av t = ⎜ x 0 x ⎟ t = ⎜
⎟( 5.00 s ) = 45.0 m
2
⎝ 2 ⎠ ⎝
⎠