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HW5_Solutions - PROBLEM 3.71 The steel plate shown will...

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Unformatted text preview: PROBLEM 3.71 The steel plate shown will support six SO-mm-diameter idler rollers mounted on the plate as shown. Two flat belts pass around the rollers, and rollers A and D will be adjusted so that the tension in each belt is 45 N. Determine (a) the resultant couple acting on the plate if a = 0.2 m, (b) the mic... value of a so that the resultant couple acting on the plate is 54 N-rn clockwise. SOLUTION . . . . (a) Note when a = 0.2 In, 1'ch 15 perpendIcular to the 1nc1n1ed 45 N forces. Have M = 2(Fd) = _(45 N)[o + 0.2 m + 2(0025 111)] _ (45 N)[Za\/§ + 2(0025 111)] For a = 0.2 In, M = —(45 N)(0.450 m + 0.61569 m) = 47.956 Nvfll or M = 48.0 N111) 4 (b) M = 54.0 N-m) tit-ruin M = Moment of couple due to horizontal forces at A and D + Moment of force—couplesystenis at C and F about C. —54.0 N-rn = —45 N[a + 0.2 m + 2(0025 m)] + [MC + MF + Fx(a + 0.2 m) + Fy(23)] where ' MC = —(45 N)(0.025 m) = -1.125 Nm MF = MC = —1.125N-m PROBLEM 3.71 CONTINUED .5 F}. =———N J27 @M —54.0 N111 = .45 N(a + 0.25 In) — 1.125 N-m —-1_125 N.m F —45 N(a + 0.2 m)— f 1.20 = a + 0.25 + 0.025 + 0.025 + % 2 3.121301r = 0.75853 :2 = 0.24303 m PROBLEM 3.73 The tension in the cable attached to the end C of an adjustable boom ABC is 1000 N. Replace the force exerted by the cable at C with an equivalent force-couple system (a) at A, (b) at B. (a) Basedcn 2F: FA = T =1000N ,. or FA =1000N “C 20‘" EMA: MA = (TsiI150°)(dA) = (1000 N)si1150°(2.25 m) = 1723.60N-m 01' MA = 1724 N-m)< (5) Based on 2F: FB = T = 1000 N or F3 =1000N “x: 2004 2MB: ME = (Tsh150°)(d3) = (1000 N)sin50°(1.25 m) = 957.56 N-m or M3 = 958N*m)4 While tapping a hole, a machinist applies the horizontal forces shown to the handle of the tap wrench. Show that these forces are equivalent to a single force, and specify, if possible, the point of application of the single force on the handle. SOLUTION Since the forces at A and B are parallel, the force at B can be replaced with the sum of two forces with one of th an opposite sense, resulting in a force-couple. the forces equal in magnitude to the force at A except wi Have FB = 26.5 N + 2.5 N, where the 26.5 N force be part of the couple. Combining the two parallel forces, Mm, = (26.5 N)[(0.030 m + 0.070 m)co525°:l = 3.60 N-rn and, M,,,,,, = 3.60N-m) H 2.50%! 1&0“er A 8| A 53!, ‘x A single equivalent force will be located in the negative z—direction. 2MB: —3.60 N-m = [(2.5 N)c0525°](a) ES” 353M _ - -, _._,i Based on a = —1.590 m F’ = (2.5 N)(cos25°i + sin 25°j) and is applied on an extension of handle BD at a distance of 1.590 In to the right of B 4 SOLUTION 3 . PROBLEM 3.96 The handpiece of a miniature industrial grinder weighs 2.4 N, and its center of gravity is located on the y axis. The head of the handpiece is offset in the x2 plane in such a way that line BC forms an angle of 25° with the x direction. Show that the weight of the handpiece and the two couples M1 and M2 can be replaced with a single equivalent force. Further assuming that MI = 0.068 N-m and M2 = 0.065 Nsrn, determine (a) the magnitude and the direction of the equivalent force, (1;) the point where its line of action intersects the 3:2 plane. First assume that the given force W and couples M1 and M2 act at the origin. NOW W = —FVj and M = M1 + M2 = —(M2coszs°)i + (M1 — M2 si1125°)k Note that since W and M are perpendicular, it follows that they can be replaced With a s1ngle equivalent force. =W or F =—W} =—(2.4N)j or F = —(2.40 N)j 4 (1:) Assume that the line of action of F passes through point P (x, 0, 2*). Then for equivalence PROBLEM 3.112 Three forces and a couple act on crank ABC. Determine the value of d so that the given system of forces is equivalent to zero at (a) point B, (2)) point D. SOLUTION Based on IF; = 0 Puma—3110: O '. Pcosa' =31b (I) and 217}. = 0 Psina — 2113 = 0 '. Psiner : 21b (2) Dividing Equation (2) by Equation (1), tano: = E 3 a: = 33.690” Substituting into Equation (1), P = i = 3.6056 lb eo333.690° or P = 3.61mi 33.7cl (61) Based on 2MB = 0 {(36056 1b)eos33.690°:|[(d + 6 in.)sin50°] —-[(3.60561b)sin33.690°][(d + 6 in.)c0550°] +(3 1b)[(6in.)sin50°] —(21b](6in.)+ 50113-561. = 0 46333.51 :- —30.286 d : 84509 in. or d = 8.45 in.‘ PROBLEM 3.112 CONTINUED (5) Based on 2MB, = 0 {(36056 1b)cos33.690”][(d + 6 in.)si1150°:| — [(36056 lb)sin33.690°:| [(d + 6 in.)cosSO° + 6 111.] +(316)[(6 1113611500] + 5016.16. = 0 —3.583 861' = 4:02.86 _,.-u-t d = 8.4509 in. or d: 8.45 in.‘ This result is expected, since R = 0 and ME = 0 for d = 8.45 in. implies that R = 0 and M = O at any ether point for the value of d found in part a. PROBLEM 3.121 The head-andsmotor assembly of a radial drill press was originally positioned with arm AB parallel to the z axis and the axis of the chuck and bit parallel to the y axis. The assemny was then rotated 25” about the y axis and 20° about the centerline of the horizontal arm AB, bringing it into the position shown. The drilling process was started by switching on the motor and rotating the handle to bring the bit into contact with the workpiece. Replace the force and couple exerted by the drill press with an equivalent force-couple system at the center 0 of the base of the vertical column. Have R = F = (44 N)[(Sil120°30525°)i — (cos 20°) j - (5111203316 25°)k] = (13.6389 N)i - (41.346 N) j — (6.3599 N)k 01' R = (13.64 Np — (41.3 N) j - (6.36 17)]; 4 Have M0 = rm. x F + MC where r3... = [(0.280 m) sin 25°]i + (0.300 m) j + [(0.280 m)cos25°]k . = (0.118333 m)1 + (0.300 m) j + (0.25377 m)k MC = (7.2 N-m)[(si1120°c0525°)i' - (66320°)j — (sin 20° sin 25°)k] = (2.2318 N-m)i — (6.7658 N—m)j - (1.04072 N-m)k 1 j k M0 = 0.118333 0.300 _0.25377 Ntrn 13.6389 411.346 -6.3599 + (2.2318i - 6.7658j —1.04072k)N-m = (10.8162 N-m)i — (2.5521 N-m) j — (10.0250 N-m)k or M0 = (10.82 N.m)i — (2.55 N-m)j — (10.03 N-m)k 4 ...
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