Recitation Week 4 Chapter 5

Transcription

1 Recitation Week 4 Chapter 5 Problem 5.5. A bag of cement whose weight is hangs in equilibrium from three wires shown in igure P5.4. wo of the wires make angles θ = 60.0 and θ = 40.0 with the horizontal. Assuming the system is in equilibrium, show that the tension in the left-hand wire is = cos θ sin(θ + θ ) () θ θ 3 Balancing forces on the cement bag, 3 =. Balancing forces on the wire joint is a bit mor complicated and deserves a free body diagram. θ + θ θ θ θ ĵ î 3 Where we ve chosen a coordinate system such that has no y component. Balancing forces in the y direction, which is what we set out to show. 0 = sin(θ + θ ) 3 cos(θ ) () cos(θ ) = 3 sin(θ + θ ) = cos(θ ) g sin(θ + θ ), (3) Problem 5.8. An object of mass m = 5.00 kg placed on a frictionless, horizontal table is connected to a string that passes over a pulley and then is fastened to a hanging object of mass = 9.00 kg as shown in igure P5.8. (a) Draw free-body diagrams of both objects. ind (b) the magnitude of that acceleration of the objects and (c) the tension in the string. m (a)

2 N m (b) Because the rope does not stretch, both objects have the same magnitude of acceleration. Using = ma on both objects, we can solve for a. = m a (4) g = a (5) g m a = a (6) (c) Plugging the solution for a back into either of the = ma equations, = m a = g = (m + )a (7) a = g 9.00 kg 9.80 m/s = m kg kg = 6.30 m/s (8) m g 5.00 kg 9.00 kg 9.80 m/s = = 3. N (9) m kg kg Problem wo objects are connected by a light string that passes over a frictionless pulley as shown in igure P5.30. Assume the incline is frictionless and take m =.00 kg, = 6.00 kg, and θ = (a) Draw free-body diagrams of both objects. ind (b) the magnitude of the acceleration of the objects, (c) the tension in the string, and (d) the speed of each object.00 s after it is released from rest. m θ (a) N m m g g (b) Because the rope does not stretch, both objects have the same magnitude of acceleration. Using = ma on both objects,

3 we can solve for a. is the heavier object, so we ll pick the positive direction to be dropping and raising m. m g = m a (0) g sin(θ) = a () = m (g + a) () g sin(θ) m (g + a) = a (3) g( sin(θ) m ) = a(m + ) (4) a = g sin(θ) m m + (5) = 9.80 m/s 6.00 kg sin(55.0 ).00 kg.00 kg kg (6) = 3.57 m/s (7) (c) Plugging the solution for a back into either of the = ma equations, ( = m (g + a) = m g + m ) sin(θ) m = m g m + + sin(θ) m m + m + (8) = m g ( + sin(θ)) = g m ( + sin(θ)) m + m + (9) = 9.80 m/s.00 kg 6.00 kg.00 kg kg ( + sin(55.0 )) = 6.7 N (0) (d) Because the string does not stretch, the speed of both objects are the same. Because the acceleration is constant, v = a t + v 0 = a t = 3.57 m/s.00 s = 7.4 m/s () Problem A car is traveling at 50.0 mi/h on a horizontal highway. (a) If the coefficient of static friction between road and tires on a rainy day is 0.00, what is the minimum distance in which the car will stop? (b) What is the stopping distance when the surface is dry and = 0.600? (a) A normal force with magnitude mg is required to keep the car from accelerating in the vertical direction (and either sinking into the pavement or levitating above it). he frictional resistance has magnitude which gives a deceleration of f = N = mg, () f = ma = mg (3) a = g. (4) Because the car is moving, you might expect the coefficient of kinetic friction would be more appropriate. However, if the wheels are not skidding (e.g. with anti-lock brakes), the tire does not slide over the road, so you use the coeffient of static friction. You would use a coefficient of kinetic friction if you were analyzing the disk/brake-pad interaction. or constant acceleration problems, v = v 0 + a(x x 0 ) (5) (see my solution to Prob..33 for a derivation). We can use this forumla to solve for the stopping distance x = x x 0 = v v 0 a = v 0 a = v 0 g. (6) Converting the initial speed to m/s, v 0 = 50.0 mi/h Plugging into our formula for stopping distance.6 km.00 mi.00 h =.4 m/s. (7) 3600 s x = v 0 g = (.4 m/s) = 55 m. (8) m/s

4 (b) Plugging the new into our formula for stopping distance which is much shorter. x = v 0 g = (.4 m/s) = 4.5 m, (9) m/s Problem wo blocks connected by a rope of negligable mass are being dragged by a horizontal force (ig. P5.47). Suppose = 68.0 N, m =.0 kg, = 8.0 kg, and the coefficient of kinetic friction between each block and the surface is (a) Draw a free-body diagram for each block. Determine (b) the acceleration of the system and (c) the tension in the rope. m m (a) N N m f f m g g (b) Because the string does not stretch, the blocks will have the same acceleration, and can be treated as a single block. µ(m + )g = (m + )a (30) a + µg = m + (3) a = µg m + (3) 68.0 N =.0 kg kg m/s =.9 m/s (33) (c) We can use the horizontal force on m to calculate the tension µm g = m a (34) m = m (µg + a) = m = m + m + (35).0 kg = 68.0 N = 7. N.0 kg kg (36) Problem A crate of wieght is pushed by a force P on a horizontal floor as shown in igure P5.63. he coefficient of static friction is, and P is directed at an angle θ below the horizontal. (a) Show that the minimum value of P that will move the crate is given by sec θ (37) tan θ (b) ind the condition on θ in terms of, for which motion of the crate is impossible for any value of P. P crate (a) he normal force must resist both the force of gravity and the vertical component of P, so N = + P sin(θ). (38)

5 his moves the crate when the horizontal component of P balances the force of friction. which is what we set out to show. P cos(θ) = N = ( + P sin(θ)) (39) P (cos(θ) sin(θ)) = (40) P ( tan(θ)) = sec(θ) (4) sec(θ) tan(θ), (4) Note that this formula is only valid when there is an actual normal force to provide friction. herefore P cos(θ) > 0. We can posit, without loss of generality, that P > 0, in which case the restriction is 90 < θ < 90. By symmetry, the situation for the backside 80 is just a mirror image of the frontside. (b) As P becomes larger, the component of our horizontal force balance becomes negligable, so we cannot move the block when P cos(θ) P sin(θ) (43) tan(θ) (44) ( ) θ arctan θ c, (45) where the last step uses the fact that tan(θ) is strictly increasing on the range θ ( 90, 90 ). What does this mean about our answer to (a)? Let s rework the condition to look more like the denominator in the (a) answer. tan(θ) (46) 0 tan(θ) (47) 0 tan(θ), (48) so the denominator is negative or zero for θ θ c. or θ just below the cutoff, the denominator is small but positive, and you get a really large value for P. or θ = θ c, the denominator is zero, and you get an infinite value for P. or θ above the cutoff, the denominator is negative, so P is also negative, which, as I pointed out in (a), is not allowed. he whole thing is a bit easier to understand if we rephrase the answer to (a) as cos(θ) sin(θ) = C cos(θ) sin(θ) = (A cos(θ) B sin(θ)), (49) where C =, A /C, and B /C = /. We can consolidate to a single trig term using Matching with our formula, sin(a ± b) = sin(a) cos(b) ± cos(a) sin(b) (50) (D sin(a b)) = (D sin(a) cos(b) D cos(a) sin(b)). (5) θ = b A = D sin(a) (53) B = D cos(a) (54) tan(a) = A B = /C /C = (55) ( ) a = arctan (56) D = B cos(a) = B ( cos arctan D (sin(a b)) = ( )) = B + µ s (5) + µ s (57) + µ = s ( ( ) ) csc arctan θ. (58)

6 his doesn t look as clean as the phrasing in (a), but it makes the dependence of P on θ much clearer. or example, P is obviously negative for θ > θ c arctan(/ ). he dependency on θ over the rest of the range is P csc(θ c θ) = sin(θ c θ) (59) Because is a positive number, / will also be positive, and θ c will be between 0 and 90. he status on all possible angles looks something like θ c cos(θ) > 0 P > 0 aking the cos(θ) > 0 portion of our P dependence (where the equation we started with in (a) applies, and combining it with the reflection (which applies when cos(θ) < 0, we get ( ( ) ) if θ c θ < 80 θ c + g csc arctan θ if 90 θ < θ c µ (60) s ( ( ) ) csc arctan 80 + θ if 80 θ c < θ 80 or 80 θ 90 which looks like + µ s θ c where I ve just plotted the θ dependence of P, setting the constant / term equal to. Note that the csc makes nice, straight lines in this polar plot.