HW04 forces: HW set 4: 6-CQ8, 6-24, 6-27, 6-34, 6-40, 6-49, 6-50, 6-57, 6-70.

Pool of problems: 6.24, 6.34, 6.49, 6.57, 6.70. Allowed equations,

Chapter 6, concept question 8: In Fig. 6-15, a horizontal force of 100 N is to be applied to a 10 kg slab that is initially stationary on a frictionless floor, to accelerate the slab, fig 6-15

A 10 kg block lies on top of the slab; the coefficient of frictionbetween the block and the slab is not known, and the block might slip. Considering that possibility, what is the possible range of values for the magnitude of the slabs accelerationaslab? (Hint: You dont need written calculations; just consider extreme values for.)

There are two possibilities corresponding to extreme values of , the static friction: (1) , and the slab pulls the block along with it, in which case 100 N is accelerating a system of mass

The second case (2) would be , and the systems acceleration is,

What is the possible range for the magnitudeablockof the blocks acceleration? In case (1), the blocks acceleration is equal to the systems acceleration, which is , and in case (2), the blocks acceleration is 0. Case 2 could be a tablecloth-slab, and a china-glass-block, just like in the movies.

Chapter 6, problem 24: A 4.50 kg block is pushed along a floor by a constant applied force that is horizontal and has a magnitude of 43.0 N. The figure gives the block's speedvversus timet as the block moves along anxaxis on the floor. The scale of the figure's vertical axis is set byvs= 5.30 m/s, so the increments are 0.530 m/s. What is the coefficient of kinetic friction between the block and the floor?

The coefficient of kinetic friction is defined as the ratio of friction-force to normal force, , in which is given by , and appears in ,

Chapter 6, problem 27: Body A weighs 102 N, and body B weighs 32 N. The coefficients of friction between A and the incline are and. The incline is . Let the positive direction of an x axis be up the incline. In unit-vector notation, what is the acceleration of A if A is initially (a) at rest, (b) moving up the incline, and (c) moving down the incline?

Before we consider any of the three cases, we write Newtons 2nd Law with appearing as a Newtons 3rd Law pair in separate force-sums (as discussed in class). Let the coefficients of static friction be , and that of kinetic friction be , in which and . The reason for this is that we dont know the direction of the friction yet, and we can multiply by +1 or -1 in order to assign the proper direction to the friction. We simply let the abovementioned be unknown symbols that run around in our calculations (as you are hopefully noting, working in symbols is like giving yourself dials to turn and tweak to whatever value you want; you can even cackle madly while doing so). We have, using (conservation of string), and using the indicated direction the problem-statement instructs us to (!),

Using the 2nd equation in , we have ; adding the 1st and 3rd equations in , the internal force/N3L-pair goes away, and we are left with an acceleration of,

This is the system-acceleration in terms of problem givens (i.e., we multiplied top and bottom by g, because we are given , etc.).

(a) Now we answer part-a. We specialize to the numbers given, for which we realize , implying , which means (i.e., the static frictions direction (but not magnitude) is known). Now we need to find out if the static friction is sufficiently strong as to hold the block A still; indeed, this happens if satisfies equation (if you compute a value of that is greater than 1, then does not satisfy the equation, and the block slides),

The block (and indeed the whole system; both blocks, that is) is, indeed, held in place.

(b) When block-A is moving up the incline, the kinetic friction opposes its motion () and the acceleration is found from effecting the replacement , yielding,

(c) If you answered (a) and (b) correctly, it is quite boring to realize, this time, the replacement in of, and we simply get,

Chapter 6, problem 34: In the figure, a slab of massm1= 24 kgrests on a frictionless floor, and a block of massm2= 9 kgrests on top of the slab. Between block and slab, the coefficient of static friction is 0.30, and the coefficient of kinetic friction is 0.20. A horizontal force of magnitude 61 N begins to pull directly on the block, as shown. In unit-vector notation, what are the resulting accelerations of(a)the block (m2) and(b)the slab (m1)?

Find out if the top block is sliding or not:

Thus, the force of static friction is overcome, and kinetic friction takes over. Then, each block accelerates at a rate given by Newtons 2nd Law, and . (a) block-2 accelerates as,

Note the negative acceleration: this is because, presumably, the right direction is +x. Sanity check:; that is, this is the case of zero block-1/block-2 friction.

(b) The 1st block accelerates as,

Chapter 6, Problem 40 SN (not on quiz): In downhill speed skiing a skier is retarded by both the air drag force on the body and the kinetic frictional force on the skis. Suppose the slope angle is, the snow is dry snow with a coefficient of kinetic friction, the mass of the skier and equipment is m, the cross-sectional area of the (tucked) skier is A, the drag coefficient is C, and the air density is. (a) What is the terminal speed? (b) If a skier can vary C by a slight amount dC by adjusting, say, the hand positions, what is the corresponding variation in the terminal speed? Give your answer in terms of the given values and g (note: dC counts as a variable, independent of C).

Terminal speed computed from condition of zero-acceleration. Thus, all forces, and their tally, appear as,

Solving for v in of , we immediately get,

Computing , we have,

Sanity check: obviously, terminal velocity should decrease if C is increased, hence the minus sign in .

Chapter 6, Problem 49: In the figure, a car is driven at constant speed over a circular hill and then into a circular valley with the same radius. At the top of the hill, the normal force on the driver from the car seat is 0. The driver's mass is 52.1 kg. What is the minimum magnitude of the normal force on the driver from the seat when the car passes through the bottom of the valley?

Intuitively, the correct answer is . The justification for this,

Chapter 6, problem 50: A 74.0 kg passenger is made to move along a circular path of radiusr= 88.0 m in uniform circular motion.(a)Figure (a) is a plot of the required magnitudeFof the net centripetal force for a range of possible values of the passenger's speedv. What is the plot's slope atv= 6.40 m/s?(b)Figure (b) is a plot ofFfor a range of possible values ofT, the period of the motion. What is the plot's slope atT= 2.40 s?

For figure a, you need force as a function of velocity, which you subsequently differentiate with respect to to compute the slope. Note the units of the slope,

For figure b, you need force as a function of period. The period enters into ,

Chapter 6, Problem 57: A puck of mass m = 1.00 kg slides in a circle of radius r = 21.0 cm on a frictionless table while attached to a hanging cylinder of mass M = 3.00 kg by a cord through a hole in the table. What speed (in m/s) keeps the cylinder at rest?

Combining the two equations , we easily eliminate , and obtain , in which ,

Chapter 6, Problem 70: The figure shows a conical pendulum, in which the bob (the small object at the lower end of the cord) moves in a horizontal circle at constant speed. (The cord sweeps out a cone as the bob rotates.) The bob has a mass of 0.018 kg, the string has length L = 0.79 m and negligible mass, and the bob follows a circular path of circumference 1.3 m. What are (a) the tension in the string and (b) the period of the motion?

Notice I drew an angle in the figure, which links many important quantities. Let be the circumference. Then, we relate to both forces and spatial dimensions (in the y and radial directions), as,

We have two instances of Newtons 2nd Law: in the y-direction, and in the radial-direction, which tell us the components of the tension,

The tension is the Pythagorean resultant of the radial and y components,

Let be the period (the letter T is already used for tension). Then, the uniform circular velocity is expressible in terms of this period, , as,

This is the period in terms of problem-givens. Plugging in numbers,