ANALYZE In all three cases (a) (c), the reading on the scale is. w = mg = (11.0 kg) (9.8 m/s 2 ) = 108 N.

Size: px

Start display at page:

Download "ANALYZE In all three cases (a) (c), the reading on the scale is. w = mg = (11.0 kg) (9.8 m/s 2 ) = 108 N."

Griselda Ford
5 years ago
Views:

1 Chapter 5 1. We are only concerned with horizontal forces in this problem (ravity plays no direct role). We take East as the +x direction and North as +y. This calculation is efficiently implemented on a vector-capable calculator, usin manitude-anle notation (with SI units understood). F a m b b b Therefore, the acceleration has a manitude of.9 m/s. 15. THINK We have a piece of salami hun to a sprin scale in various ways. The problem is to explore the concept of weiht. EXPRESS We first note that the readin on the sprin scale is proportional to the weiht of the salami. In all three cases (a) (c) depicted in Fi. 5-34, the scale is not acceleratin, which means that the two cords exert forces of equal manitude on it. The scale reads the manitude of either of these forces. In each case the tension force of the cord attached to the salami must be the same in manitude as the weiht of the salami because the salami is not acceleratin. Thus the scale readin is m, where m is the mass of the salami. ANALYZE In all three cases (a) (c), the readin on the scale is w = m = (11.0 k) (9.8 m/s ) = 108 N. LEARN The weiht of an object is measured when the object is not acceleratin vertically relative to the round. If it is, then the weiht measured is called the apparent weiht. 19. THINK In this problem we re interested in the force applied to a rocket sled to accelerate it from rest to a iven speed in a iven time interval. EXPRESS In terms of manitudes, Newton s second law is F = ma, where F = F net, a a, and m is the (always positive) mass. The manitude of the acceleration can be found usin constant acceleration kinematics (Table -1). Solvin v = v 0 + at for the case where it starts from rest, we have a = v/t (which we interpret in terms of manitudes, makin specification of coordinate directions unnecessary). Thus, the required force is F ma mv / t. ANALYZE Expressin the velocity in SI units as 195

2 196 CHAPTER 5 v = (1600 km/h) (1000 m/km)/(3600 s/h) = 444 m/s, we find the force to be F v 444m s 5 m 500 k N. t 1.8s LEARN From the expression F mv /, t we see that the shorter the time to attain a iven speed, the reater the force required. 6. Some assumptions (not so much for realism but rather in the interest of usin the iven information efficiently) are needed in this calculation: we assume the fishin line and the path of the salmon are horizontal. Thus, the weiht of the fish contributes only (via Eq. 5-1) to information about its mass (m = W/ = 8.7 k). Our +x axis is in the direction of the salmon s velocity (away from the fisherman), so that its acceleration ( deceleration ) is neative-valued and the force of tension is in the x direction: T T. We use Eq. -16 and SI units (notin that v = 0). v (.8 m/s) x 0.11 m m/s v v a x a Assumin there are no sinificant horizontal forces other than the tension, Eq. 5-1 leads to T ma T 8. 7 k 36m s which results in T = N. b c h 9. We choose up as the +y direction, so a ( 3.00 m/s )j (which, without the unitvector, we denote as a since this is a 1-dimensional problem in which Table -1 applies). From Eq. 5-1, we obtain the firefihter s mass: m = W/ = 7.7 k. ˆ. (a) We denote the force exerted by the pole on the firefihter F f p 5-1. Since Fnet ma, we have F and apply Eq. fp ĵ F F ma F fp fp 71 N (7.7 k)( 3.00 m/s ) which yields F fp = 494 N. (b) The fact that the result is positive means F fp points up.

3 197 (c) Newton s third law indicates Ff p Fpf Fpf 494 N., which leads to the conclusion that (d) The direction of Fpf is down. 37. (a) Since friction is neliible the force of the irl is the only horizontal force on the sled. The vertical forces (the force of ravity and the normal force of the ice) sum to zero. The acceleration of the sled is F 5. N as 0. 6 m s. m 8.4 k s (b) Accordin to Newton s third law, the force of the sled on the irl is also 5. N. Her acceleration is F 5. N a 013. m s. m 40k (c) The accelerations of the sled and irl are in opposite directions. Assumin the irl 1 starts at the oriin and moves in the +x direction, her coordinate is iven by x a t. The sled starts at x 0 = 15 m and moves in the x direction. Its coordinate is iven by 1 xs x ast. They meet when x xs, or at x0 ast. This occurs at time t By then, the irl has one the distance x 1 xa x 0. a a s 15 m0.13 m/s 0 at a a s 0.13 m/s 0.6 m/s.6 m. 51. The free-body diarams for m1 and m are shown in the fiures below. The only forces on the blocks are the upward tension T and the downward ravitational forces F1 m1 and F m. Applyin Newton s second law, we obtain:

4 198 CHAPTER 5 T m m a 1 1 m T m a which can be solved to yield m m 1 a m m1 Substitutin the result back, we have T mm 1 m m 1 (a) With m1 1.3 k and m.8 k, the acceleration becomes.80 k 1.30 k (9.80 m/s ) 3.59 m/s 3.6 m/s. a.80 k 1.30 k (b) Similarly, the tension in the cord is (1.30 k)(.80 k) (9.80 m/s ) 17.4 N 17 N. T 1.30 k.80 k 61. THINK As more mass is thrown out of the hot-air balloon, its upward acceleration increases. EXPRESS The forces on the balloon are the force of ravity m (down) and the force of the air F a (up). We take the +y to be up, and use a to mean the manitude of the acceleration. When the mass is M (before the ballast is thrown out) the acceleration is downward and Newton s second law is M Fa Ma After the ballast is thrown out, the mass is M m (where m is the mass of the ballast) and the acceleration is now upward. Newton s second law leads to F a (M m) = (M m)a. Combin the two equations allows us to solve for m.

5 199 ANALYZE The first equation ives F a = M( a), and this plus into the new equation to ive M b Ma a b M m b M m a m. a LEARN More enerally, if a ballast mass m is tossed, the resultin acceleration is a which is related to m via: a a m M, a showin that the more mass thrown out, the reater is the upward acceleration. For a a, we et m Ma /( a), which arees with what was found above. 80. We take down to be the +y direction. (a) The first diaram (shown below left) is the free-body diaram for the person and parachute, considered as a sinle object with a mass of 80 k k = 85 k. F a is the force of the air on the parachute and m is the force of ravity. Application of Newton s second law produces m F a = ma, where a is the acceleration. Solvin for F a we find F m a 85 k 9.8 m/s.5 m/s 60 N. a (b) The second diaram (above riht) is the free-body diaram for the parachute alone. F a is the force of the air, m p is the force of ravity, and Fp is the force of the person. Now, Newton s second law leads to Solvin for F p, we obtain m p + F p F a = m p a. F m a F 5.0 k.5 m/s 9.8 m/s 60 N 580 N. p p a 85. (a) Since the performer s weiht is (5 k)(9.8 m/s ) = 510 N, the rope breaks.

6 00 CHAPTER 5 (b) Settin T = 45 N in Newton s second law (with +y upward) leads to which yields a = 1.6 m/s. T m T ma a m

2. We apply Newton s second law (Eq. 5-1 or, equivalently, Eq. 5-2). The net force applied on the chopping block is F F F

2. We apply Newton s second law (Eq. 5-1 or, equivalently, Eq. 5-2). The net force applied on the chopping block is F F F Chapter 5 1. We are only concerned with horizontal forces in this problem (gravity plays no direct role). We take East as the +x direction and North as +y. This calculation is efficiently implemented on