Problem#1
A woman at an airport is towing her 20.0-kg suitcase at constant speed by pulling on a strap at an angle θ above the horizontal (Fig. 1). She pulls on the strap with a 35.0-N force, and the friction force on the suitcase is 20.0 N. Draw a free-body diagram of the suitcase. (a) What angle does the strap make with the horizontal? (b) What normal force does the ground exert on the suitcase?
 |
| Fig.1 |
Answer:
msuitcase = 20.0 kg, F = 35.0 N
∑Fx = max
–20.0 N + F cos θ = 0
∑Fy = may
+n + F sin θ – mg = 0
 |
| Fig.2 |
(a) angle does the strap make with the horizontal is
F cos θ = 20.0 N
cos θ = 20.0 N/35.0 N = 0.571
θ = 55.20
(b) n = mg – F sin θ
n = 196 N – 35.0(0.821) N = 167 N
Problem#2
A 3.00-kg block starts from rest at the top of a 30.0° incline and slides a distance of 2.00 m down the incline in 1.50 s. Find (a) the magnitude of the acceleration of the block, (b) the coefficient of kinetic friction between block and plane, (c) the friction force acting on the block, and (d) the speed of the block after it has slid 2.00 m.
Answer:
Given: m = 3.00 kg, θ = 30.00, x = 2.00 m and t = 1.50 s, then
 |
| Fig.3 |
(a) the magnitude of the acceleration of the block is
x = ½ at2
2.00 m = ½ a(1.50 s)2
a = 1.78 m/s2
(b) the coefficient of kinetic friction between block and plane.
Along x:
∑Fx = ma
0 – f + mg sin θ = ma
f = m(g sin θ – a) (1)
Along y:
∑Fy = ma
n – mg cos θ = 0
n = mg cos θ (2)
so that
µk = f/n = (g sin θ – a)/g cos θ
µk = tan θ – a sec θ/g
µk = tan 30.00 – (1.78 m/s2) sec 30.00/(9.80 m/s2) = 0.368
(c) the friction force acting on the block is
f = m(g sin θ – a)
f = 3.00 kg(9.80 m/s2 sin 30.00 – 1.78 m/s2) = 9.37 N
(d) the speed of the block after it has slid 2.00 m, we use
vf2 = vi2 + 2aΔx
vf2 = 0 + 2(1.78 m/s2)(2.00 m)
vf = 2.67 m/s
Problem#3
A Chevrolet Corvette convertible can brake to a stop from a speed of 60.0 mi/h in a distance of 123 ft on a level roadway. What is its stopping distance on a roadway sloping downward at an angle of 10.0°?
Answer:
 |
| Fig.4 |
First we find the coefficient of friction:
∑Fy = 0
+n – mg = 0; then n = mg and
∑Fx = ma
–f = ma
–µsmg = ma, then a = –µsg
So that, by using
vf2 = vi2 + 2aΔx
0 = vi2 + 2(–µsg)Δx
µs = vi2/2gΔx
µs = (88 ft/s)2/[2(32.1 ft/s2)(123 ft)] = 0.981
Now on the slope
∑Fy = 0
+n – mg cos 100 = 0; then n = mg cos 100 and
∑Fx = ma
–f + mg sin 100 = ma
–µsmg cos 100 + mg sin 100 = ma
Then
a = –µsg cos 100 + g sin 100
So that, by using
vf2 = vi2 + 2aΔx
0 = vi2 + 2(–µsg cos 100 + g sin 100)Δx
Δx = vi2/[2g(–µs cos 100 + sin 100)]
Δx = (88 ft/s)2/[2(32.1 ft/s2)(–0.981 cos 100 + sin 100]
Δx = 152 ft
Problem#4
A 9.00-kg hanging weight is connected by a string over a pulley to a 5.00-kg block that is sliding on a flat table (Fig.5). If the coefficient of kinetic friction is 0.200, find the tension in the string.
 |
| Fig.5 |
Answer:
First, consider the block moving along the horizontal. The only
force in the direction of movement is T. Thus, ∑F = max.
 |
| Fig.6 |
T – fk = (5 kg)a
With
fk = µkn = µkm1g = 0.200(5.0 kg)(9.80 m/s2) = 9.80 N, then
T – 9.80 = (5 kg)a (1)
Next consider the block that moves vertically. The forces on it are
the tension T and its weight, 88.2 N.
We have ∑Fy = ma
88.2 – T = (9.0 kg)a (2)
Note that both blocks must have the same magnitude of acceleration. Equations (1) and (2) can be
added to give
78.4 N = (14 kg)a
a = 5.60 m/s2 and
∴ T = 9.80 N + 5 kg x 5.60 m/s2 = 37.8 N
Problem#5
Three objects are connected on the table as shown in Figure 7. The table is rough and has a coefficient of kinetic friction of 0.350. The objects have masses of 4.00 kg, 1.00 kg, and 2.00 kg, as shown, and the pulleys are frictionless. Draw free-body diagrams of each of the objects. (a) Determine the acceleration of each object and their directions. (b) Determine the tensions in the two cords.
 |
| Fig.7 |
Answer:
For m1:
∑Fy = m1a
m1g – T12 = m1a
(4.00 kg)(9.80 m/s2) – T12 = (4.00 kg)a
39.2 N – T12 = (4.00 kg)a
 |
| Fig.8 |
For m2:
∑Fx = m2a
T12 – f = m2a
T12 – 0.350(1.00 kg)(9.80 m/s2) = (1.00 kg)a
T12 – 3.43 N = (1.00 kg)a
For m3:
∑Fy = m3a
T23 – m3g = m3a
T23 – (2.00 kg)(9.80 m/s2) = (2.00 kg)a
T23 – 19.6 N = (2.00 kg)a
add up the equations (1), (2) and (3), we get
+39.2 N – 3.43 N – 19.6 N = (7.00 kg)a
a = 2.31 m/s2 down for m1 left for m2 and up for m3.
(b) Now 39.2 N – T12 = (4.00 kg)(2.31 m/s2)
T12 = 30.0 N
And T23 – 19.6 N = (2.00 kg)(2.31 m/s2)
T23 = 24.2 N
Post a Comment for "Forces of Friction Problems and Solutions 2"