✔ 最佳答案
23. (a) Let F be the force exerted by the triceps muscle.Take moment about the humerus,15g x 35cos(30) = F x 2.5cos(30) + 2g x 15cos(30)where g is the acceleration due to gravityi.e. F = 1942 N = 1.942 kN(b) It is already given in the question that the force is vertically upward( c) For vertical equilibrium, 15g + 1942 = 2g + F’where F’ is the force given by the hermerushence, F’ = 2070 N = 2.1 kN(d) It is clear that F’ must be downward for equilibrium to be maintained. 31. For vertical equilibrium, the y-component of the force on each hinge= 27g/2 N = 132 NThe coordinate of the centre of mass of the door is at (0.455 m, 1.05 m)Take moment about the lower hinge,27g x 0.455 = (Fx).(0.9 – 0.3 – 0.3)where Fx is the force along the x-axis at the upper hinge and ((0.9 – 0.3 – 0.3) m is the distance from the lower hinge.i.e. Fx = 80 N in the –x direction in order to provide a counterclockwise momentFor equilibrium in the horizontal direction, the horizontal force at the lower hinge must be at the +x direction. Therefore the two forces are (-80)i + (132)j N and (80i + 132j) N 35. Sorry, the graphs are not clear, can’t help 37. Let F be the reaction at the roller, which makes an angle of 20 degrees with the horizontalLet R be the normal reaction at the bottom of the plankHence, frictional force = uR, where u is the coefficient of frictionFor horizontal equilibrium: F.cos(20) = uR ------------- (1)for vertical equilibrium: 445 = R + F.sin(20) -------------- (2)Take moment about the bottom of the plank,445 x 3.05cos(70) = F.(3.05/sin(70))hence, F = 143 NFrom (2), R = 396.1 NHence, from (1) u = 143.cos(20)/396.1 = 0.34
2011-10-25 17:42:12 補充：
41. The angle at which the ladder makes with the ground = arc-cos(0.762/2.44) = 71.8 degrees
Consider the right hand portion of the ladder, take moment about the top of ladder
2011-10-25 17:43:01 補充：
R.(2.44cos(71.8)) = F.(2.44/2).sin(71.8) ------------ (1)
Where R is the normal reaction on the right foot of the ladder
F is the tension in the tie-rod
Consider the left hand portion of the ladder, take moment about the top of ladder
2011-10-25 17:43:30 補充：
854 x (2.44-1.8).cos(71.8) + F.(2.44/2).sin(71.8) = R’(2.44cos(71.8))
where R’ is the normal reaction at the left foot of the ladder
But for vertical equilibrium, R + R’ = 854 --------- (3)
Hence, 854 x (2.44-1.8).cos(71.8) + F.(2.44/2).sin(71.8) = (854 – R) x (2.44cos(71.8)) ----- (2)
2011-10-25 17:43:39 補充：
Solving (1) and (2) for R and F gives F = 211 N, and R = 320 N
Hence from (3) R’ = 534 N
2011-10-25 20:46:06 補充：
Because of words limit here...Q41 is given in "opinion (意見) section"
