Problem#1
Two blocks connected by a rope of negligible mass are being dragged by a horizontal force F (Fig. 1). Supposethat F = 68.0 N, m1 = 12.0 kg, m2 = 18.0 kg, and the coefficient of kinetic friction between each block and the surface is 0.100. (a) Draw a free-body diagram for each block. (b) Determine the tension T and the magnitude of the acceleration of the system. |
| Fig.1 |
(a) See Figure to the right
Block 1:
T - µm1g = m1a
T – 0.100(12.0 kg)(9.80 m/s2) = (12.0 kg)a
T – 11.76 N = (12.0 kg)a (1)
Block 2:
F – T - µm2g = m2a
68.0 N – T – 0.100(18.0 kg)(9.80 m/s2) = (18.0 kg)a
68.0 N – T – 17.64 N = (18.0 kg)a (2)
Add up the equations (1) and (2), we get
38.6 N = (30 kg)a
a = 1.29 m/s2
and
T – 11.76 N = (12.0 kg)(1.29 m/s2)
T = 27.2 N
Problem#2
A block of mass 3.00 kg is pushed up against a wall by a force P that makes a 50.0° angle with the horizontal as shown in Figure 3. The coefficient of static friction between the block and the wall is 0.250. Determine the possible values for the magnitude of P that allow the block to remain stationary.
 |
| Fig.3 |
(Case 1, impending upward motion) Setting
∑Fx = 0
P cos 50.00 – n = 0
n = P cos 50.00
fs,max = µsn = µs P cos 50.00
fs,max = 0.250(P)cos 50.00 = 0.161P
Setting
∑Fy = 0
P sin 50.00 – fs, max – mg = 0
P sin 50.00 – 0.161P – (3.00 kg)(9.80 m/s2) = 0
Pmax = 48.6 N
 |
| Fig.4 |
Setting
∑Fy = 0
P sin 50.00 + fs, max – mg = 0
P sin 50.00 + 0.161P – (3.00 kg)(9.80 m/s2) = 0
Pmax = 31.7 N
Problem#3
You and your friend go sledding. Out of curiosity, you measure the constant angle θ that the snow-covered slope makes with the horizontal. Next, you use the following method to determine the coefficient of friction µk between the snow and the sled. You give the sled a quick push up so that it will slide up the slope away from you. You wait for it to slide back down, timing the motion. It turns out that the sled takes twice as long to slide down as it does to reach the top point in the round trip. In terms of θ, what is the coefficient of friction?
Answer:
 |
| Fig.5 |
When the sled is sliding uphill
∑Fy = 0
+n – mg cos θ = 0
n = mg cos θ
then
fk = µkn = µk mg cos θ
∑Fx = ma
mg sin θ + µkmg cos θ = maup
aup = g(sin θ + µk cos θ)
so that
Δx = ½ auptup2
When the sled is sliding down, the direction of the friction force is reversed:
mg sin θ - µkmg cos θ = madown
adown = g(sin θ - µk cos θ)
Δx = ½ adowntdown2
Now, tdown = 2tup
½ auptup2 = ½ adown(2tup) 2
aup = 4adown
g(sin θ + µk cos θ) = 4g(sin θ - µk cos θ)
5µk cos θ = 3 sin θ
µk = 0.6 tan θ
Problem#4
The board sandwiched between two other boards in Figure 6 weighs 95.5 N. If the coefficient of friction between the boards is 0.663, what must be the magnitude of the compression forces (assume horizontal) acting on both sides of the center board to keep it from slipping?
 |
| Fig.6 |
Answer:
Since the board is ini equilibrium, ∑Fx = 0 and we see that the normal forces must be the same on both sides of the board. Also, if the minimum normal forces (compression forces) are being applied, the board is on the verge of slipping and friction force on each side is
f = fs,max = µsn
The board is also in equilibrium in the vertical direction, so
∑Fy = 2f – mg = 0, orf = mg/2
The minimum compression force needed is then
n = f/µs
n = mg/2µs
n = 95.5 N/(2 x 0.663) = 72.0 N
Problem#5
A block weighing 75.0 N rests on a plane inclined at 25.0° to the horizontal. A force F is applied to the object at 40.0° to the horizontal, pushing it upward on the plane. The coefficients of static and kinetic friction between the block and the plane are, respectively, 0.363 and 0.156. (a) What is the minimum value of F that will prevent the block from slipping down the plane? (b) What is the minimum value of F that will start the block moving up the plane? (c) What value of F will move the block up the plane with constant velocity?
Answer:
(a) the minimum value of F that will prevent the block from slipping down the plane is
∑Fy = 0
n + F sin 150 – 75 N cos 250 = 0
n = 75 N cos 250 – F sin 150 = 67.97 – 0.259F
fs,max = µsn
fs,max = 0.363(67.97 – 0.259F)
fs,max = 24.67 – 0.094F
For equilibrium:
∑Fx = 0
F cos 150 + fs,max – 75 sin 150 = 0
F cos 150 + 24.67 – 0.094F – 75 sin 150 = 0
F = 8.05 N
(b) ∑Fx = 0
F cos 150 – fs,max – 75 sin 150 = 0
F cos 150 – 24.67 + 0.094F – 75 sin 150 = 0
F = 53.2 N
(c) fk = µkn = 10.6 – 0.040F
Since the velocity is constant, the net –force is zero:
F cos 150 – 10.6 + 0.040F – 75 sin 150 = 0
This gives
F = 42.0 N
Problem#6
Review problem. One side of the roof of a building slopes up at 37.0°. A student throws a Frisbee onto the roof. It strikes with a speed of 15.0 m/s and does not bounce, but slides straight up the incline. The coefficient of kinetic friction between the plastic and the roof is 0.400. The Frisbee slides 10.0 m up the roof to its peak, where it goes into free fall, following a parabolic trajectory with negligible air resistance. Determine the maximum height the Frisbee reaches above the point where it struck the roof.
Answer:
We must consider separately the disk when it is in contact with the roof and when it has gone over the top into free fall. In the first case, we take x and y as parallel and perpendicular to the surface of the roof:
∑Fy = 0
+n – mg cos θ = 0
n = mg cos θ
then friction is
fk = µkn = µkmg cos θ
∑Fx = ma
–fk – mg sin θ = ma
a = –g(sin θ + µk cos θ)
a = –(9.80 m/s2)(sin 370 – 0.400 cos 370) = –9.03 m/s2
The Frisbee goes ballistic with speed given by
vxf2 = vxi2 + 2aΔx
vxf2 = (15 m/s)2 + 2(–9.03 m/s2)(15 m – 0)
vxf = 6.67 m/s
For the free fall, we take x and y horizontal and vertical:
Vyf2 = vyi2 + 2aΔy
0 = (6.67m/s sin 370)2 + 2(–9.80 m/s2)(yf – 10 m sin 370)
yf = 6.02 m + (4.01 m/s)2/(19.6 m/s2) = 6.84 m
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