Problem#1
A van accelerates down a hill (Fig. 1), going from rest to 30.0 m/s in 6.00 s. During the acceleration, a toy (m = 0.100 kg) hangs by a string from the van’s ceiling. The acceleration is such that the string remains perpendicular to the ceiling. Determine (a) the angle θ and (b) the tension in the string.
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| Fig.1 |
Answer:
Choose the x-axis pointing down the slope.
vf = vi + at
30.0 m/s = 0 + a(6.00 s)
a = 5.00 m/s2
Consider forces on the toy
∑Fx = max
mg sin θ = ma
sin θ = a/g = (5.00 m/s2)/(9.80 m/s2) = 0.5102
θ = sin-1(0.5102) = 30.70
and
∑Fy = may
T – mg cos θ = 0
T = (0.100 kg)(9.80 m/s2) cos 30.70 = 0.843 N
Problem#2
In Figure 3 the incline has mass M and is fastened to the stationary horizontal tabletop. The block of mass m is placed near the bottom of the incline and is released with a quick push that sets it sliding upward. It stops near the top of the incline, as shown in the figure, and then slides down again, always without friction. Find the force that the tabletop exerts on the incline throughout this motion.
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| Fig.3 |
Answer:
Throughout its up and down motion after release the block has
∑Fy = may
+n – mg cos θ = 0
n = mg cos θ
Let R = Rxi + Ryj, represent the force of table on incline. We have
∑Fx = max
+Rx – n sin θ = 0
Rx = mg cosθ sinθ = ½ mg sin2θ
∑Fy = may
–Mg – n cos θ + Ry = 0
Ry = Mg + mg cos2θ
So that
R = (½ mg sin2θ)i + g(M + mcos2θ)j
Problem#3
A magician pulls a tablecloth from under a 200-g mug located 30.0 cm from the edge of the cloth. The cloth exerts a friction force of 0.100 N on the mug, and the cloth is pulled with a constant acceleration of 3.00 m/s2. How far does the mug move relative to the horizontal tabletop before the cloth is completely out from under it? Note that the cloth must move more than 30 cm relative to the tabletop during the process.
Answer:
Take +x in the direction of motion of the tablecloth. For the mug:
∑Fx = ma
0.1 N = (0.2 kg)a
a = 0.5 m/s2
Relative to the tablecloth, the acceleration of the mug is
0.5 m/s2 – 3.0 m/s2 = –2.5 m/s2
The mug reaches the edge of the tablecloth after time given by
Δx = vxt + ½ axt2
–0.3 m = 0 + ½ (–2.5 m/s2)t2
t = 0.490 s
The motion of the mug relative to tabletop is over distance
Δx = ½ axt2 = ½ (0.5 m/s2)(0.490 s)2 = 0.0600 m = 6.00 cm
The tablecloth slides 30 cm + 6 cm = 36 cm over the table in this process.
Problem#4
A mobile is formed by supporting four metal butterflies of equal mass m from a string of length L. The points of support are evenly spaced a distance θ apart as shown in Figure 5. The string forms an angle θ1 with the ceiling at each end point. The center section of string is horizontal. (a) Find the tension in each section of string in terms of θ1, m, and g. (b) Find the angle θ2, in terms of θ1, that the sections of string between the outside butterflies and the inside butterflies form with the horizontal. (c) Show that the distance D between the end points of the string is
D = L(2 cos θ1 + 2 cos [tan-1 (½ tan θ1)] + 1)
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| Fig.5 |
Answer:
(a) Apply Newton’s second law to two points where butterflies are attached on either half of mobile (other half the same, by symmetry)
T2 cos θ2 – T1 cos θ1 = 0 (1)
T1 sin θ1 – T2 sin θ2 – mg = 0 (2)
T2 cos θ2 – T3 = 0 (3)
T2 sin θ2 – mg = 0 (4)
Substituting (4) into (2) for T2 sin θ2,
T1 sin θ1 – mg – mg = 0
T1 = 2mg/sinθ1
Substitute (3) into (1) for T2 cosθ2:
T3 – T1 cosθ1 = 0
T3 = T1 cosθ1
Then
T3 = (2mg/sinθ1)cosθ1
T3 = 2mg cotθ1
From Equation (4),
T2 = mg/sinθ2
(b) Divide (4) by (3):
T2 sinθ2/T2 cosθ2 = mg/T3
tan θ2 = mgtanθ1/2mg = ½ tanθ1
then
θ2 = tan-1(½ tan θ1)
so that
T2 = mg/{sin[tan-1(½ tan θ1)}
(c) D is the horizontal distance between the points at which the two ends of the string are attached to the ceiling.
D = 2lcos θ1 + 2l cosθ2 + l
D = l{2cos θ1 + 2cosθ2 + 1}
D = L{2cos θ1 + 2cos[tan-1(½ tan θ1)] + 1}/5
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