220A Solutions. Assignment 5.

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Maurice Stokes
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1 4-40. Draw the free-bod diagram. 220A Solutions Assignment 5. - cos sin here are two stumbling points. he first is to resist the temptation to draw a force pointing down the inclined plane. he onl forces are the normal force that the plane eerts on the block and the pull of gravit. he second is to get the angle straight. Do it carefull once and ou'll have no further trouble with it. One wa is as follows: is perpendicular to the inlclined plane and is perpendicular to the floor. he angle between these two perpendiculars and the angle between the inclined plane and the floor is the same. Another good wa is to draw it carefull. here are onl two angles in the triangle and one look tells us which is correct and whitch is incorrect. correct inc he net force is

2 hus, ewton's 2nd law gives us ΣF = sin ΣF = - cos ΣF = sin = ma ΣF = - cos = ma but, a = 0 ( it isn't even moving in the -direction), so = cos from the -direction. From the -direction, sin = ma. so g sin = a We could have looked this up in Eample 4-7, p. 95, and plugged in the numbers and have finished the problem in one minute. hat's oka after we have gone through it carefull once or twice and we understand it throughl, but we can never learn phsics b searching around for a formula that works. We must learn to reason from the basic principles or we largel waste our time studing phsics. a = 9.8 m/s 2 sin 22 = 3.67 m/s 2. he acceleration is constant so we are permitted to use the constant acceleration formulas. ime is not mentioned so we'll use v2 = v a ( - 0 ) which gives v 2 = 0 +2(3.67 m/s 2 )(2 m - 0) = 88. m 2 /s 2, so v = ± 9.39 m/s. he block is moving in the plus direction, so v = m/s. How long does it take to descend? 4-4. We know the acceleration from the previous problem a = 3.67 m/s 2. his is a practical problem of the tpe we did in the previous chapter, so we need to figure out what we know. We know the initial velocit ( - 4 m/s) and the final velocit (zero, since it stops). We're asked for the distance so we'll use v2 = v a ( - 0 ) which gives so 0 2 = (- 4.0 m/s) 2 + 2(3.67 m/s 2 )( - 0), = - (- 4.0 m/s)2/2(3.67 m/s 2 ) = m. We have begun with the origin at the bottom of the plane, so the final value of is negative; that is to sa, up the plane.

3 - 2.7 m We can get the time to ascend from v = v 0 + at which gives or 0 = - 4 m/s m/s t t = (4 m/s)/3.67 m/s =.09 s. he same time is required to descend, so the total time is twice this, 2.8 s Draw the free-bod diagrams. ote that we have chosen to be positive down for block 2. his is not necessar, just convenient because the positive -direction for block is the positive -direction for block 2. m g 2 m g 2 For block, ΣF = - m g, but since the acceleration is zero in the - direction, the 2nd law immediatel gives = m g. ΣF =, so the 2nd law gives = m a ()

4 For block 2, ΣF = - + m 2 g, (remembering that is positive down), so the 2nd law gives - + m 2 g = m 2 a 2 (2) he accelerations are written a and a 2 respectivel, but these are numbericall equal so we write them a = a 2 = a, and eqs and 2 become two equations in the two unknowns and a. Eplain to a doubting friend that a = a 2. Eplain wh it was convenient to choose positive down for block 2; i.e., what would have happened if twe had chosen it positive up. Solve the two equations two unknowns anwa ou want; for eample, we could add the two and the drops out giving m 2 g = m a + m 2 a, so a = m 2 g/(m + m 2 ) putting this back into eq gives = m m 2 g/(m + m 2 ) Draw the free-bod diagrams. ote that we have used ewton's third law in using the same force on each of the two blocks. Also, since the masses are the same in this problem, both are written m. a b 2 -m gcos30 c 30 m g m g 2 30 m g m gsin30 he net force on block 2 is ΣF = -. It is much simpler to use an inclined coordinate sstem for block. here are two trick parts. First, we need to get the angle straight to see that the angle between and m g is 30. he second trick part is to resolve the weight into components in the inclined coordinate sstem which is detailed in c. he -component of the weight is sin30 and the - component is - m g cos30.

5 he net force on block is ΣF = - cos and ΣF = - + m gcos. hus, ewton's 2nd law gives us ΣF = - + m g sin = m a ΣF = - m g cos = m a but, a = 0 ( it isn't even moving in the -direction), so = m gcos from the -direction. From the -direction, - + m g sin = m a () he net force on block 2 is ΣF = - m 2 g hus, ewton's 2nd law gives us - m 2 g = m 2 a (2) he acceleration of block in the -direction is numbericall equal to the acceleration of block 2 in the -direction, so we write a for block equal to a and a for block 2 equal to a. Eplain this an a wa that a student who does not understand it will get it. If ou have trouble with this, be sure to see our instructor. hus, we have two equations in two unknowns which we ma solve b adding eqs and 2 to eliminate the which ields m g sin - m 2 g = (m + m 2 )a a = (m g sin - m 2 g)/(m + m 2 ) (3) a = ( kg 9.8 m/s 2 sin30 - kg 9.8 m/s 2 )/( kg + kg) = m/s 2 he minus sign means that block accelerates up the plane and block 2, down. (b) If the sstem remains at rest, then a = 0 and eq 3 becomes 0 = (m g sin - m 2 g)/(m + m 2 ), so m g sin = m 2 g, or m 2 = m sin = kg sin 30 = 0.5 kg. (c) Use either eq or 2 to get ; for eample, from eq 2

6 = m 2 (a + g) = kg ( m/s m/s 2 ) = 7.35 for case (a), and = m 2 (a + g) = 0.5 kg ( m/s 2 ) = 4.9 for case (b) Draw the free-bod diagram. If ou are still having trouble with the angle and the resolution into components, see problem cos sin m g he net force in the -direction is zero, so = cos. In the - direction, ΣF = sin so the 2nd law gives sin = ma, thus a = g sin = 9.8 m/s 2 sin9.5 =.62 m/s 2. he acceleration is constant and we are given the distance and asked for the time. hus, we ma use = 0 + v 0 t + 2 at2, which gives 3 m = 0 + v 0 t m/s2 t 2, so t2 = 3 m/ 2.62 m/s2 = 3.70 s 2, thus t = ±.92 s, from which the phsical answer is t = +.92 s. he acceleration does not depend on the mass, so the answer is the same (a) Draw the free-bod diagram.

7 F f he Earth eerts the force on the bo of magnitude straight down, the plane eerts a force perpendicular to the plane with magnitude, and the plane eerts a frictional force parallel to the plane with magnitude F f. If there is an confusion about which wa the frictional force should point, note that the net force on the bo must be zero since it is not accelerated. umbericall F f = sin. (b) If the bo is sliding down the the plane, nothing changes qualitativel in the free-bod diagram. he problem does not specif if the bo is accelerated or not. If it descends at constant speed, F f = sin ; if it speeds up, then F f < sin ; and if it slows down, F f > sin. hus, the length of the arrow in the figure changes according to the situation, but the qualitative diagram is the same in all cases. (c) If the bo slide up the plane, then the frictional force is reversed. F f 5-5. Draw the free-bod diagram.

8 F f he diagram is drawn assuming the acceleration of the train to be to the right. he net force is ΣF = F f ΣF = -, so the 2nd law is F f = ma () - = ma but a = 0, so = (2) It's important to get the reasoning straight. he acceleration in eq is our acceleration. In order for ou not to slip, our acceleration must be equal to that of the train; a = 0.20g It is the frictional force force F f that provides our acceleration, according to eq. We can get the maimum frictional force, because we know the normal force in eq 2; thus, F f < µ s. Substituting F f = µ s into eq will lead to the minimum value of µ s and using eq 2 µ s = ma µ s = ma. he mass divides out so our mass doesn't matter. µ s g = a, so µ s = a /g = 0.2 g/g = Draw the free-bod diagrams. ote that we have chosen to be positive down for block 2, so that the accelerations of the two blocks are the same.

9 F f m g 2 m g 2 For block, ΣF = - m g, but since the acceleration is zero in the - direction, the 2nd law immediatel gives = m g. his means that the maimum frictional force is F f = µ s = µ s m g () ΣF = - F f, so the 2nd law gives - F f = m a (2) so, combining eqs and 2, - µ s m g = m a (3) For block 2, ΣF = - + m 2 g, (remembering that is positive down), so the 2nd law gives - + m 2 g = m 2 a (4) ote that the two accelerations are the same. Equations 3 and 4 become two equations in the two unknowns and a. Solve the two equations two unknowns anwa ou want; for eample, we could add the two and the drops out giving - µ s m g + m 2 g = m a + m 2 a, so a = (- µ s m g + m 2 g)/(m + m 2 ). hus a = 0 results from (- µ s m g + m 2 g) = 0 so m =m 2 /µ s = 2 kg/(0.4) = 5 kg.

10 If the sstem begins to move, µ k becomes the coefficient of friction. he free-bod diagram is the same and the acceleration is again zero, so everthing is the same ecept µ k replaces µ s. m =m 2 /µ s = 2 kg/(0.3) = 6.67 kg.

Lab 5 Forces Part 1. Physics 211 Lab. You will be using Newton s 2 nd Law to help you examine the nature of these forces.

Lab 5 Forces Part 1. Physics 211 Lab. You will be using Newton s 2 nd Law to help you examine the nature of these forces. b Lab 5 Forces Part 1 Phsics 211 Lab Introduction This is the first week of a two part lab that deals with forces and related concepts. A force is a push or a pull on an object that can be caused b a variet