PH 1A - Fall 2017 Solution 2

Transcription

1 PH A - Fall 207 Solution Let the tension of the cable between be T AB. First, we analyze the total torque τ on the boom with respect to fulcrum C. It should be 0 since the boon is static. τ = T AB L sin (lb) L sin (lb) L 2 sin 60 = 0 We then get T AB = 2200 lb. For the force acting on the boom at point C by the hinge, we should use force equilibrium condition on the boom. x-direction: = T AB cos 30 = lb y-direction: F C x F C y = 2000(lb) + 400(lb) T AB sin 30 = +300 lb Similarly, the force equilibrium condition at B gives x-direction: = T AB cos 30 = lb y-direction: F B y F B x = 2000(lb) T AB sin 30 = +900 lb where F B is the force acting on the cable by the boom. 8.8 Let the tension of the cable be T. We focus on object B first. Since it moves with constant speed, the net force of B has to be 0. Therefore, we have: T = m B g ()

2 Next, A moves with constant speed, so its net force is also 0. The equation of force equilibrium can be written as where θ = 30 and µ = 0.2 in this problem. Combine () and (2), we have m A g sin θ T µm A g cos θ = 0 (2) m B = m A (sin cos 30 ) = = 2.6 (kg) m A (3) QP4 (a) The free body diagrams are Fig.. The normal forces N ij are the force of the i th block on the j th block, or just N i for the force of the ground on block i. N 3 T N 32 T N 4 M N 3 M 2 N 3 T M T 3 2 T M 4 θ θ T 2 F g F g F g N 23 F g Figure : QP4.a (b) Newton s Second Law for m and m 2 : m a X = N 3 m 2 a 2X = T m a Y = T m g m 2 a 2Y = N 32 m 2 g 2

3 (c) Since there is no motion of m 2 relative to m 3, we must have a 2X = a 3X. Also, we know a Y = 0. Thus: T = m g a 3X = a 2X = T m 2 = m m 2 g (d) The whole system moves with the same horizontal acceleration, a = a 3X. The only external force is F, so we have: F = m sys a sys F = (m + m 2 + m 3 + m 4 ) m m 2 g (e) Newton s Second Law for m 4 yields: m 4 a 4X = m 4 a = T 2 cos θ T 2 = m 4a cos θ m 4 a 4Y = 0 = N 4 m 4 g T 2 sin θ N 4 = m 4 g + T 2 sin θ ( N 4 = m 4 (g + a tan θ) = m 4 g + m ) tan θ m 2 (f) Since pulley is massless, the net force is zero. See Fig. 2. T F p T Figure 2: QP4.f Clearly, F P = T (ˆx + ŷ). So F P = 2T = 2m g, and it is directed at φ = tan [F P Y /F P X ] = 45. 3

4 QP (a) If you make the reasonable assumption that the parameters of the problem indicate that m 2 will move to the right (up the inclined plane), then the solution is as follows. (See Fig. 3.) Both blocks move uphill: N m N m 2 f 2 F g f f2 g2 N Figure 3: QP.a In the y-direction: N = m g cos θ F f = µ N N 2 = N + m 2 g cos θ = (m + m 2 )g cos θ and in the x-direction: F f2 = µ 2 N 2 F f m g sin θ = m a F F f F f2 m 2 g sin θ = m 2 a 2 The condition that m does not slide relative to m 2 implies that a = a 2 = a. We want to explore the maximum value of a at which m still sticks. F f s,max m g sin θ > m a Thus F f s,max = µ s m g cos θ a < (µ s cos θ sin θ)g 4

5 When a reaches that limit, m will slip off. For certain values of the parameters, the F required to cross that threshold, which we shall call F crit is actually insufficient to start moving either block at all, i.e. if µ 2s is sufficiently greater than µ 2k. Thus there is an additional requirement.... Treating m and m 2 as one object for the time being Thus F (m + m 2 )g sin θ F f2 > 0 F f2 = µ 2s (m + m 2 )g cos θ F up > (m + m 2 )g(sin θ + µ 2s cos θ) is the force necessary for m 2 to slip, and for anything to happen at all. (b) Now for part (b) we are expected to actually calculate F crit for the acceleration condition from part (a). F < m 2 a crit + m 2 g sin θ + F f + F f2 substituting in a crit = (µ s cos θ sin θ)g from above plus F f and F f2 we get F crit = m 2 (µ s cos θ sin θ)g + m 2 g sin θ + µ s m g cos θ + µ 2k (m + m 2 )g cos θ canceling and combining terms we get F crit = (m + m 2 )g(µ s + µ 2k ) cos θ The total answer to part b) is then F max = max(f crit, F up ) in order that both conditions be met. QP20 (Thanks to Prof. Michael Cross) The question is about a rope of length 2l draped over a frictionless nail. If one side is of length l + x and the other l x, the longer side will pull the shorter side over the nail, and the speed of the motion v = ẋ will increase. 5

6 We have an intuitive picture that the force along the rope in the direction of the motion on the longer side is larger than the force along the rope in the direction of the motion on the shorter side, so there is a net force along the direction of motion causing the acceleration of the whole rope. I think the questioner was asking for this result in part (a) (which was only worth point and so an easy answer must be expected). This is not, however, the net force in any sense of the addition of vector forces on the rope, which would be the sum of the gravitational forces downwards on the two parts of the string, and the normal reaction force upwards of the nail on the rope. This latter version of the net force is quite hard to calculate and basically needs a solution of the whole problem. Here are the two approaches to the question. Intuitive Approach (a) The net force pulling the rope along the motion (down on the long side, up on the short side) is F = ρg(l + x) ρg(l x) = 2ρgx. (b) The total mass being accelerated is 2ρl, so the equation of motion is 2ρl dv Using the hint given, this can be written as which integrates to dv = 2ρgx = g x. (4) l v dv dx = d(v 2 ) 2 dx = (g/l)x, v 2 = (g/l)x 2 + c. The initial condition is v = 0 for x = 0, so that c = 0, giving v 2 = (g/l)x 2. (5) This intuitive approach can be put on a firm foundation using the ideas of generalized coordinates and forces, and d Alembert s Principle. You will learn about these ideas if you take a more advanced mechanics class such as Ph06. If you want an inkling of these ideas, you can find discussions on Wikipedia. 6

7 Sticking to the wording of the question To find the net force on the rope in the sense of gravity downwards minus force from the nail upwards, and to solve for the motion in the conventional way using Newton s second law of motion, we need to split the system into pieces: I ll use the three fixed boxes outlined in the figure (I ve made the nail larger for display purposes.) The free body diagrams for the two pieces of the rope are shown below. The motion is quite easily found using the conservation of energy (kinetic plus potential), which you will cover in a later lecture. 7

8 Consider box A. We use force = rate of change of momentum. However, the argument is a little tricky because the momentum is changing for two reasons: due to the forces and due to the fact the momentum is flowing into the box because extra rope is being added at the rate of a length v per unit time, with a momentum ρv per unit length. To confirm that this momentum flux must be included in the force equation, David Politzer suggested the following simpler situation: First consider a single, open railroad car (on a level, frictionless track). There are no horizontal forces on the car. So its horizontal velocity is constant. Now, coal is dropped onto the car from above. If the coal drops vertically (in the track frame), the car s speed will decrease keeping the total horizontal momentum a constant. However, if the coal is dropped from a chute that gives each piece a horizontal component of velocity equal to the car s velocity, the car velocity remains constant while its momentum and mass (including the new, loaded coal) increase. Thus the correct equation is d(mv) = ρg(l + x) T + ρv 2. Since dm/ = ρv, this simplifies to ρ(l + x) dv = ρg(l + x) T, (6) which is just force = mass times acceleration. You can also derive this expression by going to the non-inertial frame moving with the string and then equating the total force, including the inertial force from the acceleration, to zero, since there is no acceleration in this frame. The corresponding equations for box B are and then d(mv) ρ(l x) dv = ρg(l x) + T ρv 2 = ρg(l x) + T. (7) 8

9 Adding the two equations gives 2ρl dv = 2ρgx. This is the same as Eq. (4) and the argument proceeds from there. This answers part (b). The net force on the rope can be obtained form the total rate of change of momentum for the whole system. This gives a net force in the downwards direction of strength F = d [ρ(l + x)v ρ(l x)v] = 2ρ d (xv) = 4ρv2 = 4ρg x 2, l where the explicit solution (5) is used to get the last two equalities. This answers part (a) of the question. Alternatively, this result is given by considering box C. Again we use force = rate of change of momentum and again there is a flow of momentum into the box down momentum is flowing out on the right-hand side, and up momentum is flowing in on the right-hand side. This gives N = 2T 2ρv 2. Calculate T by subtracting Eq. (6) from Eq. (7) T = ρgl ρx dv = ρgl ρv2 where the explicit solution (5) is used in the last equality. Thus the normal force is N = 2ρgl 4ρv 2. Finally, this gives the net force downwards on the rope to be F = 2ρgl N = 4ρv 2 = (4ρg/l)x 2. 9

10 Additional subtlety (Thanks to Prof. Harvey Newman) The rope actually comes off of the nail before the end passes over the nail, because the centrifugal force of the rope wrapping around the nail becomes larger than the gravitational force holding the rope on the nail. This happens when T = ρv 2, giving x = l/ 2 and v = gl/2. 0