HW 3 Help 3. ORGANIZE AND PLAN We can use Newton s second law to find the acceleration of the rocet. The force on the rocet will be the vector sum of all the forces acting on it, which are the force due to gravity (straight down) and the force from the rocet engine (straight up). We will use a coordinate system where ĵ represents the upward vertical direction. Known: m = 3.5 g; F engine = 95.3 N ˆj SOLVE The force acting on the rocet is [Eq. ] F =F +mg= 95.3 N j 3.5 g 9.8 m/s ( j ) = 6 N j. engine ˆ ˆ ˆ From Newton s second law, the acceleration of the toy rocet is [Eq. ] a=f / m= 6 N j /3.5 g = 7.4 m/s j. ˆ REFLECT The rocet is accelerating upward at almost twice the acceleration due to gravity. This is a significant acceleration, compared to what we experience on a daily basis. To appreciate the acceleration due to gravity, try to catch a stic that falls from a stationary position, with your hand initially positioned above the stic. It can be done, but it s not easy. Imagine now the rocet accelerating at almost twice that rate ˆ 35.ORGANIZE AND PLAN vector form, F=ma. This problem involves the application of Newton s second law in Known: m = 00 g = 0. g; a 0.55 m/s i ˆ 0.650 m/s ˆj SOLVE Inserting the nown quantities (with the correct units) into Newton s second law, we have [Eq. ] ˆ ˆ ˆ ˆ F ma 0. g 0.55 m/s i 0.650 m/s j 0.055 N i 0.0650 N j This is the force needed to generate the desired acceleration. The direction of this force is [Eq. ] 0.055 atan 68.6,.4. 0.0650 From Eq. () we now the force vector is in the second quadrant, so we the direction of the force must be.4 above the horizontal. The magnitude of the force is [Eq. 3] F 0.55 N 0.0650 N 0.0698 N REFLECT Note that we had to convert the mass from grams to ilograms so that the result of our calculation would be in SI units (i.e., Newtons).
The crane supplies an upward force on the beam, and gravity supplies a downward force. The force will be the vector sum of the two. Once we find the force, we can use Newton s second law to find the acceleration of the beam. We chose a coordinate system for this problem in which ĵ is oriented vertically upward. Known: m = 85 g; F 960 N ˆj ; g 9.8 m /s ˆ j 44.ORGANIZE AND PLAN crane SOLVE The force is the vector sum of all the forces on the beam, which in this case is gravity and the force applied by the crane [Eq. ]: F F mg 960 N j (85 g)(9.8 m/s ) j 47 N j crane ˆ ˆ ˆ We insert this result into Newton s second law to find the acceleration of the beam [Eq. ]: F ma, a F / m j0.8 m/s j (85 g) (47 N) ˆ ˆ REFLECT Most of the force applied by the crane is used to overcome gravity. If the force (i.e., crane + gravity) is applied to the beam for sec, the beam will gain a speed of 0.8 m/s in the upward direction. How much force must the crane now exert to maintain this speed constant? Simply enough to mae F. From Eq. this is F mg. 0 crane Mae a drawing of the situation on which all the forces for each case are drawn. Sum up the forces and use Newton s second law to find the normal force in each case. 46.ORGANIZE AND PLAN N N 5 N N N N mg mg mg 5 N N (a) (b) (c) Known: m =.5 g SOLVE From Newton s second law it is nown that, since the boo is stationary, the force must be zero. From the figure above, the force for case (a) is [Eq. ] Thus the normal is N( ˆj ). F 0 nmg j n mg j j j ˆ ˆ (.5 g)(9.8 m/s ) ˆ Nˆ
Repeating the same calculation for case (b) gives [Eq. ] F 0 nmg j 5 N j n mg j j j j j ˆ ˆ ˆ 5 N ˆ N ˆ 5 N ˆ 7 N ˆ Thus the normal force applied by the table on the boo has been reduced to For case (c), we have [Eq. 3] F 0 nmg j 5 N j n mg j j j j j ˆ ˆ ˆ 5 N ˆ N ˆ 5 N ˆ 37 N ˆ ˆ 7 ( ). N j Thus, when you push down on the boo with 5 N of force, the table must now supply a force of 37 N( ˆj ) on the boo. REFLECT What is the normal force supplied by the floor on each of the 4 table legs for each situation, if we assume the boo is positioned at the center of the table? You can find this by doing the same calculations, but replacing the mass m by the mass of the table plus the boo. Also, don t forget that the floor is pushing up on all 4 legs equally, so the force applied on each leg will be ¼ of the total normal force. This problem involves a frictionless pulley and a presumably massless wire, so the force applied by the wire on each hanging bloc is the same. In addition, the acceleration of m will be of equal magnitude but antiparallel to the acceleration of m [Eq. ]: a a. Known: m m SOLVE (a) The force diagram is shown in the figure below. 6.ORGANIZE AND PLAN T T N m m m g m g (b) Using Newton s second law, the acceleration on m is [Eq. ] and on m it is [Eq. 3] F ma T mg ma Tmg j 3 ˆ F ma T mg ma Tmg j Using Eqs. and 3 to express a and T in terms of the other variables in Eq., we get [Eq. 4] ˆ
From Eq. [Eq. 5], (c) If 0.50 g ˆ ma m mg ˆ m mg ˆ m m m ma ma mgmg j a j j m m g a m m m and m, then [Eq. 6] 0.00 g a.40 m/s a.40 m/s ˆj ˆj ˆj REFLECT Consider what would happen if m 0. In this case, m would accelerate down with the acceleration due to gravity, and m would accelerate upward with the same rate. The opposite would be true if m 0. Since the tightrope-waler does not accelerate, the force on her must be zero (Newton s second law). Known: m 63 g; 9.5 SOLVE (a) The force diagram is shown in the figure below. 68.ORGANIZE AND PLAN N u u T T (b) Summing the forces in the vertical direction, we have [Eq. ] F 0mgTsin ˆj mg 63 g9.8 m/s ˆj T sin9.5 T 870 N REFLECT The horizontal component of the rope tension is T cos 845 N, and the vertical component is T sin 309 N. 4
7.ORGANIZE AND PLAN Using Eq. 4.6, the magnitude of the iic friction force is [Eq. ] f n. The friction force will act to oppose the velocity, which we assume is in the î direction. Therefore [Eq. ], ˆ find the acceleration, and then the distance traveled. Known: = 0.03; x x 6 m iˆ ; v 0 m/ s; 0 f mg i. From this and Newton s second law we can f 0 0 ˆ v v i SOLVE Using Newton s second law and Eq., the acceleration of the puc is [Eq. 3] F f ma ˆ 0.039.8 m/s ˆ 0.3 m/s ˆ mg a i i i m (b) To find the distance traveled before stopping, we use [Eq. 4] vf v0 at and [Eq. 5] x x v t at to find [Eq. 6] f 0 0 / v0 v0 xf x0 v0 /a a a v x x a i 6 m 0.3 m/s i 4.0 m/s i ˆ ˆ ˆ 4.0 m/si. 0 f 0 Thus, the puc s initial velocity must be REFLECT Notice that we have to be careful with the direction of the vector quantities in Eq. 6 to ensure the correct result. We can use the same force diagram as for Problem 75. If the car is moving down the incline at a constant speed, then by Newton s second law there is no force on the car. Known:.4 ; a 0 m/s SOLVE Using Newton s second law and Eq. 4.6, we get [Eq. ] 76.ORGANIZE AND PLAN F f mg i ma ˆ sin ˆ sinˆ0 ˆ ˆ n i mg i mgcos i mgsin i 0 tan 0.04 REFLECT This coefficient of iic friction is between that of the wooden bloc (0.87, Problem 75) and that of the stone on ice (0.04, Problem 73). 5
94. ORGANIZE AND PLAN The horizontal force on the car generates the centripetal acceleration (see Eq. 4.9). Known: F 790 N i ˆ ; R 0 m; v.5 m/s ˆ SOLVE r (a) See the force diagram below. Car N v F r R cos(u) View from above R =. m N F r u mg View from in front of car There are two diagrams, the upper one is the view from above the car, the lower one is the view from in front of the car. (b) Using Eq. 4.9, we find the mass of the car is [Eq. ] mv Fr R FR 790 N 0 m r m 54 g v.5 m/s REFLECT Notice that the units cancel properly in Eq.. 6
09.ORGANIZE AND PLAN Use the inematic equations (see, e.g., Problem 4, Eq. 3) to calculate the acceleration. Use Newton s second law to calculate the force due to iic friction from the acceleration. Known: m.90 g; xx0 3.5 m; v0.0 m/s SOLVE From, e.g., Problem 4, Eq. 3, we calculate the magnitude of the acceleration [Eq. ]. v.0 m/s 0 a 0.678 m/s x x0 3.5 m Using Newton s second law and Eq. 4.6, we can find the coefficient of iic friction [Eq. ]. F ma n ma mg ma 0.678 m/s a/ g 0.069 9.8 m/s REFLECT The attentive reader might have noticed a sign difference between Eq. 3, Problem 4, and Eq. of this problem. The difference is that the former is a vector equation, while the later relationship is scalar. Since the acceleration in this problem is due to the iic friction force, the direction of the acceleration vector must be antiparallel to the velocity vector. 7