HW5 - ENGINEERING MECHANICS - Dynamics W.F. Riley &...

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Unformatted text preview: ENGINEERING MECHANICS - Dynamics W.F. Riley & L.D. Sturges 15—4* A lS-kg block of ice slides on a horizontal surface for 20 m before it stops. If the initial speed of the block was 15 m/s, determine a. The force of friction between the block of ice and the surface. b. The kinetic coefficient of friction Mk between the block of ice and the surface. Solution The free-body diagram includes the weight of the block W = 15(9.81) = 147.15 N, the normal force N and friction force F exerted on the block by the surface. The equations of motion are +4 SE; = max: -F = 15a +T 2F =ma: N—147.15=O Y Y ‘ where ay = 0 since there is no motion in the vertical direction. § Therefore N = 147.15 N 15a = -F Rewriting the x-component of acceleration using the chain-rule of differentiation dv dv dx dv a=—=——-—-—=V dt dx dt dx and integrating 15V dv -F dx 2 15V /2 -Fx + C = 1687.5 — Fx where the constant of integration has been chosen so that v = 15 m/s when x = 0. Then, if v = 0 when x = 20 m, F = 84.4 N ............................................. Ans. and the kinetic coefficient of friction is u F/N = 0.573 ............... . . . . . . . . ................. Ans. k ENGINEERING MECHANICS - Dynamics W.F. Riley & L.D. Sturges 15-25* Two blocks A and B are connected by a flexible cable as shown. The kinetic coefficient of friction between block A and the inclined surface is Mk = 0.15 and the system is released from rest. If the weight of body A is 50 lb and block B strikes the horizontal surface 3 s after being released, determine a. The acceleration of body 8. b. The weight of body B. c. The tension in the cable while the blocks are in motion. Solution Since the two blocks move in separate directions, separate free-body diagrams must be drawn for each block. Using coordinates along and normal to the surface for block A, the equations of motion are 50 2 = : — — ‘ o = — +& F mAa T BA 50 Sin 40 32.2 aA (a) 0 + E = : - = 0 5 FAY mAaAy NA 50 cos 40 (b) +T Z = : - . = Fay mBaBy T 32 2mB mBaB -rn (c) where the y—component of acceleration of block A is zero since it does not move in the direction normal to the surface, only the y—component of motion gives useful information for block B, and a = —a we: "13% A B since both aA and a3 are positive upward. Therefore N = 38.3022 lb = constant F = 0.15NA = 5.7453 lb = constant (Problem 15—25 continues ...) velocity and VB YB so that VB block B hits 0 H and subtracting Eq. a from Eq. ENGINEERING MECHANICS - Dynamics 5.7453 + 32.1394 - 32.2m c gives = mBa — 1.5528(—aB) B 37.8847 - 32.2m B — constant + . 5 8 m3 1 5 2 position gives a t + C B l 2 + aBt /2 Integrating the (constant) acceleration to get the = a t ft/s B C2 = O and yB = 10 the floor (yB aB(3)2/2 + 10 2 —2.2222 ft/s 1.3789 slug 32.2mB 41.3 lb 44.4 2 + aBt /2 10 ft where the constants of integration have been chosen ft when t = 0. Then, if = 0) when t = 3 s, 2 = 2.22 ft/s ¢ . . . . . . . . . . . . . . . . . . .. Ans lb . . . . . . . . . . . . . . . . . . . . . . . . . . . . .. Ans . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .. Ans W.F. Riley & L.D. Sturges (Problem 15-25 — cont.) ‘ ENGINEERING MECHANICS - Dynamics W.F. Riley & L.D. Sturges 15—34 A 55-kg skydiver jumps from a stationary balloon at a height of 3000 m. The drag force exerted on her body by the air when she is in a spread-eagle position can be expressed as FD = 0.180v2 where F is in Newtons and V is in meters per second. Determine her terminal velocity and the time required to reach 95 percent of her terminal velocity. Solution The free-body diagram of the skydiver's body includes her weight W and the air drag Fb. Only the yhcomponent of the equations of motion gives useful information 2 +T 2F = ma : 0.180V — 55(9.8l) = 55 y y where y, v, and a are all positive upward. Rewriting the acceleration 2 305.556a = V - 2997.5 and integrating to get the velocity gives I 205.556 dV I dt V - 2997.5 1 5.5810 tanh (v/54.75) = t + c = t where the constant of integration is zero since the skydiver starts from rest. Rewriting gives v = 54.75 tanh (t/5.5810) m/s and as t 6 m V‘9 = 54.8 m/s ................................... Ans. Vterm The skydiver will reach 95 percent of the terminal velocity when v = 0.95(54.75) = 54.75 tanh (t/5.5810) t = 10.22 s ........................................ Ans. ENGINEERING MECHANICS — Dynamics W.F. Riley & L.D. Sturges 15—54* A 100-g apple falls from a tree branch 3 m above the ground. As it falls, a strong cross wind exerts a constant horizontal force of 0.05 N on the apple. Determine the horizontal distance traveled by the apple as it falls to the ground. Solution The free-body diagram of the apple includes its weight W and the wind force. The equations of motion are +9 2F ma : 0.05 x x +T 2F ma : —0.1(9.81) y y Therefore, the x— and y- components of acceleration are 2 2 a = 0.5 m/s a = -9.81 m/s X Y and they are both constant. Integrating the (constant) accelerations to get the velocity and position of the apple gives V 0.5t m/s v —9.81t m/s X Y 2 2 x 0.25t m y 3 - 4.905t m where the constants of integration have been chosen so that 3 3 m when t = 0. The apple will hit the ground when 2 y 3 — 4.905t = 0 t 0.78206 s at which time its x-position is x = 0.1529 m ENGINEERING MECHANICS - Dynamics W.F. Riley & L.D. Sturges 15-57 A circular disk rotates in a horizontal plane. A.3—1b block rests on the disk 8 in. from the axis of rotation. The static coefficient of friction between the block and the disk is 0.50. If the disk starts from rest with a constant angular acceleration of 0.5 rad/sz, determine the length of time required for the block to begin to slip. Solution The free-body diagram of the block includes its weight an a normal force t N exerted on the block by the surface, and a friction force exerted on the block by the surface. The friction force is represented in terms of its n- and t-components. The equations of motion are 3 «+2 = : =— Fn man Fn 32.2 m 3 2 + EFt - mat. Ft - 32.2 rw +T 2F = ma : N — 3 = 0 z z where the z—component of acceleration is zero since the block has no vertical motion, 2 a = 0.5 rad/s = constant w = 0.5t rad/s Therefore N = 3 lb , . (3,J¢ mew) F; = 0.03106 lb 2 Ft = 0.01553t lb 2 2 2 2 2 2 F = (0.03106) + (0.01553t ) S (0.50N) = (1.5 lb) and the block will begin to slip when t = 9.83 s ........................................ Ans. ENGINEERING MECHANICS - Dynamics 15-69 A small (W = 2 lb) chunk of ice is initially at rest at the very top of the hemispherical observatory. If a gust of wind gives the ice an initial speed of v = l ft/s, determine the 0 angle 9 where the ice will lose contact with the roof. Assume that friction between the ice and the roof of the observatory is negligible. Solution The free—body diagram of the ice includes its weight W and the normal force N exerted on it by the cylindrical roof. Using components along and normal to the surface, the equations of motion are +2 ZFt = mat: mg sin 9 = my +Y 2F = ma : mg cos 9 - N = n n Rewriting the tangential component 2 mv /R W.F. Riley & L.D. of acceleration and integrating to get the velocity of the ice as a function of 9 I vR d9 = I gR sin 6 d9 I V dv 2 V /2 —gR cos 9 + C = 0.5 + gR(l — cos 6) where the constant of integration has been chosen so that 0 V = 1 ft/s when 6 = 0 . Then, N = mg cos 6 — m[ :0 + 29(1 - cos 6] and the ice will leave the surface when A7: 0 6 = 48.100 the normal force is Sturges ...
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HW5 - ENGINEERING MECHANICS - Dynamics W.F. Riley &...

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