'- '- p = 800(i cos j sin 270 ) = Oi - -'-,--,--J~36. T = ITI(icos 22 + j sin 22 ) = ITI(0.927i

Transcription

1 Problem 3.6 A zoologist estimates that the jaw of a predator, Martes, is subjected to a force P as large as 800 N. What forces T and M must be exerted by the temporalis and masseter muscles to support this value of P? ' -'-,--,--J~36 '- '- Solution: Resolve the forces into scalar components, and solve the equilibiium equations... Express the forces in terms of horizontal and vertical unit vectors: T = ITI(icos 22 + j sin 22 ) = ITI(0.927i j) p = 800(i cos j sin 270 ) = Oi - 800j = IMj(icos j sin 144 ) = IMI( i j) Iy the equilibrium conditions, =O=T+M+P=O like terms: LFx = (0.927ITI Ml)i = 0 (I) = (0.375ITI MI - 800)j = 0 (2) tbe first equation, ITI = (0.809) IMI = MI 'tute tho ~s value IIlto the second equation, reduce algebraically, solve: IMj = 874 N, ITI = NJ

2 Problem 3.17 In Problem 3.16, determine the magnitude of the total friction force exerted on the tow truck's tires. (This is the friction force the truck's tires must exert to prevent the truck and car from sliding down the slope.) Solution: Use the value of the tension from the previous problem L F'\ : F kn sin 10 - T cos 8 = kn Solving: I F = 9.43 kn I 100 F T

3 Problem 3.22 A construction worker holds a 180-kg crate in the position shown. What force must she exert on the cable? Solution: Eqns. of Equilibrium: LF)' = T I COS 5 - T 2 sin 30 - mg = 0 { LFx mg = T2cos30 (180)(9.81) - NTI sins" = 0 Solving, we get T1 = 1867 N T2 = 188 N 5 -y x mg = ( 180) (9.81) N

4 Problem 3.30 An astronaut candidate conducts experiments on an airbearing platform. While he carries out calibrations, the platform is held in place by the horizontal tethers AB, AC, and AD. The forces exerted by the tethers are the only horizontal forces acting on the platform. If the tension in tether AC is 2 N, what are the tensions in the other two tethers? Solution: Isolate the platform. The angles a and fj are Also, tan a = - = 0.429, ( ) tan fj = - = 0.857, (3.0) 3.5 fj = The angle between the tether AB and the positive x axis is (180 - fj), hence B -I~ 1.5m ; I 3rm~D...L 1~3rii)-I-t?- I I B TAB = ITABI(i cos(l80 - fj) + j sin( fj» TAB = ITABI(-icosfJ+jsinfJ). D The angle between The tension is the tether AC and the positive x axis is (180 + a). TAC = ITAcl (i cos(l80 + a) + j sin( a» = ITAcl(-icosa-jsina). Solve: The tether AD is aligned with the positive x axis, TAD = ITADli + OJ. The equilibrium condition: ITADI = (ITAcl Sin sin(a fj + fj»). For ITACI = 2 N, a = 23.2 and fj = 40.6, Substitute and collect like terms, ITABI = 1.21 N, ITADI = 2.76 N L Fy = (ITABIsin fj - ITAcl sin a)j = O.

5 Problem 3.51 The cable AB is 0.5 m in length. The unstretched length of the spring is 0.4 m. When the 50-kg mass is suspended at B, the length of the spring increases to 0.45 m. What is the spring constant k? Solution: The Geomelry Law of Cosines and Law of Sines 0.7 m F 1 : TAB sin e + F sin </> N = N

6 Problem 3.SS The mass of each pulley of the system is m and the mass of the suspended object A is ma. Determine the force T necessary for the system to be in equilibrium. Solution: Draw free body diagrams of each pulley and the object A. Each pulley and the object A must be in equilibrium. The weights of the pulleys and object A are W = mg and W A = mag. The equilibrium equations for the weight A, the lower pulley, second pulley, third pulley, and the top pulley are, respectively, B - W A = 0, 2C - B W = 0, 2D - C - W = 0, 2T - D - W = 0, and Fs - 2T - W = o. Begin with the first equation and solve for B, substitute for B in the second equation and solve for C, substitute for C in the third equation and solve for D, and substitute for D in the fourth equation and solve for T, to get T in terms of Wand W A. The result is I IT WA 3W WA 7W 0= and T = ' 8 8' or in terms of the masses, T = ~(ma + 7m).

7 Problem 3.88 The cable AB is 0.5 m in length and the ul1stretched length of the spring Be is 0.4 m. The spring constant k is 5200 N/m. When the 50-kg mass is suspended at B, what is the resulting length of the stretched spring? Solution: Introduce the distances band h. Then we have 4 unknowns (F, TAB, b, h, LBc). We have the constraint and equilibrium equations 0.5 In = J b2 + h2 LBe = J (0.7 In - W + h2 F = (5200 N/m)(LBe m) B '\' ~Fx: ---TAB b 0.7 m - b 0.5 m LBe + ---F=O F '\' h h ~ F y : -0- TAB + - F N = 0.5 m LBe Solving we find h = m, b = 0.37\ m, LBe = m, F = 364 N, TAB = 343 N GOBe = m I N

8 Problem 3.92* The cable AB keeps the 8-kg collar A in place on the smooth bar CD. The y axis points upward. Determine the distance s from C to the collar A for which the tension in the cable is 150 N m Y x Z Solution: From the figure, the coordinates of the points (in meters) are B(O, 0.5, 0.15), C(O.4, 0.3, 0), and D(0.2, 0, 0.25). The first unit vector is of the form, «XI - xk)i + (YI - YK)j - (ZI - ZK )k) e'k = ---;============== J (XI - XK)2 + (YI - YK)2 + (ZI - ZK )2 ' where I K takes on the value CD. The coordinates of point A are given by Ax = Cx + secdx, Ay = Cy + secoy, where we do not know the value of s. The equations of equilibrium for this problem are: LFy = TABeABy + FNy - W = 0, where TAB = 150 N. The weight of the collar is given by, Y 0.15 m I ~I 1_0.4 m--t 1 B(.).! AB //"~ 'r >,,\ /' 'I~C" 0.5 m I F N ~ '. ~ s 0.3 m,./,/ 1:---.-,'- D -c:/ 0.25-,r- m - J_x "" j,_ -L.- z/--0.2j~/ T 180 I n B _120 N \ '-- =:::: \\" ~ " \ Distance. s (m) W = mg, or W = (8)(9.81) = N. The condition that the force FN is perpendicular to CD is We have three equilibrium equations plus the dot product equation in the four unknowns, s and the three components of F N. Several methods of solution are open to us. Any iterative algebraic solution method should give the result s = m and that Alternalive Solulion: The complication in the algebra in the solution is because we do not know the location of point A. We can assume the location of A is known (assume that we know the distance.1') and solve for the value of the tension in cable AB which corresponds 10 that location for A. We can plot the value of the tension versus the distance s and find the value of s at which the tension is 150 N. If we do this, we get the plot shown. From the plot, s ~ N. FN = 80.7i j k N.