Problem#1
An object of mass 5.00 kg, attached to a spring scale, rests on a frictionless, horizontal surface as in Figure 1. The spring scale, attached to the front end of a boxcar, has a constant reading of 18.0 N when the car is in motion. (a) If the spring scale reads zero when the car is at rest, determine the acceleration of the car. (b) What constant reading will the spring scale show if the car moves with
constant velocity? (c) Describe the forces on the object as observed by someone in the car and by someone at rest outside the car.
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| Fig.1 |
Answer:
(a) If the spring scale reads zero when the car is at rest, the acceleration of the car is given by
∑Fx = ma
T = ma
a = T/m = 18.0 N/5.00 kg = 3.60 m/s2 to thr right
(b) if v = constant, then a = 0. So, T = 0 (This is also an equilibrium situation.)
(c) Someone in the car (noninertial observer) claims that the forces on the mass along x are T and a fictitious force (–ma). Someone at rest outside the car (inertial observer) claims that T is the only force on M in the x-direction.
Problem#2
If the coefficient of static friction between your coffee cup and the horizontal dashboard of your car is µs = 0.800, how fast can you drive on a horizontal roadway around a right turn of radius 30.0 m before the cup starts to slide? If you go too fast, in what direction will the cup slide relative to the dashboard?
Answer:
We adopt the view of an inertial observer. If it is on the verge of sliding, the cup is moving on a circle with its centripetal acceleration caused by friction
∑Fy = may
+n – mg = 0
n = mg
and
∑Fx = mv2/r
f = mv2/r
µsmg = mv2/r
v = (µsgr)1/2 = [(0.80)(9.80 m/s2)(30.0 m)]1/2 = 15.3 m/s
If you go too fast the cup will begin sliding straight across the dashboard to the left.
Problem#3
A 0.500-kg object is suspended from the ceiling of an accelerating boxcar as in Figure 3. If a = 3.00 m/s2, find (a) the angle that the string makes with the vertical and (b) the tension in the string.
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| Fig.3 |
Answer:
The only forces acting on the suspended object are the force of gravity mg and the force of tension T, as shown in the free-body diagram. Applying Newton’s second law in the x and y directions,
∑Fx = ma; T sin θ = ma (*) and
∑Fy = 0; T cosθ – mg = 0 or T cosθ = mg (**)
(a) Dividing equation (*) by (**) gives
tan θ = a/g
θ = tan-1(3.00/9.80) = 17.00
(b) From Equation (1),
T = ma/sinθ
T = (0.50 kg)(3.00 m/s2)/sin 17.00 = 5.12 N
Problem#4
A small container of water is placed on a carousel inside a microwave oven, at a radius of 12.0 cm from the center. The turntable rotates steadily, turning through one revolution in each 7.25 s. What angle does the water surface make with the horizontal?
Answer:
Given: r = 12.0 cm = 0.120 m, T = 7.25 s,
The water moves at speed, given by
v = 2πr/T
v = 2π(0.120 m)/(7.25 s) = 0.104 m/s
The top layer of water feels a downward force of gravity mg and an outward fictitious force in the turntable frame of reference, with acceleration is
a = v2/r = (0.104 m/s)2/(0.120 m) = 0.0901 m/s2
It behaves as if it were stationary in a gravity field pointing downward and outward at
θ = tan-1(a/g)
θ = tan-1(0.0901/9.80) = 0.5270
Its surface slopes upward toward the outside, making this angle with the horizontal.
Problem#5
A person stands on a scale in an elevator. As the elevator starts, the scale has a constant reading of 591 N. As the elevator later stops, the scale reading is 391 N. Assume the magnitude of the acceleration is the same during starting and stopping, and determine (a) the weight of the person,
(b) the person’s mass, and (c) the acceleration of the elevator.
Answer:
(a) Fmax = mg + ma = 591 N (*)and
Fmin = mg – ma = 391 N (**)
the weight of the person, adding equation (*) by (**) gives
2mg = 982 N
mg = 491 N
(b) the person’s mass is, m = 491 N/(9.80 m/s2) = 50.1 kg
(c) the acceleration of the elevator, subtracting equation (*) by (**) gives
2ma = 200 N
a = 100 N/(50.1 kg) = 2.00 m/s2
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