皮皮-
更新1:
for 35
the left graph is T against (y/L)
and the slope is +600 the right graph is Fh against (y/L)
and the slope is -300
更新2:
and 1 more question: A crate
in the form of a cube with edge lengths of 1 .2 m
contains a piece of machin ery; the center of mass of the crate and its contents is located 0.30 m above the crate"s geometrical center.
更新3:
The crate rests on a ramp that makes an angle 0 with the honzontal. As 0 is increased from zero
an angle will be reached at which the crate will either tip over or start to slide down the ramp. If the coefflcient of static friction g
beeen ramp and crate is 0.60
更新4:
(a) does the crate tip or slide and (b) at what angle 0 does this occur? If p
: 0.70
(c) does the crate tip or slide and (d) at what angle g does this occur? (Hint: At the onset of tipping
where is the normal force located?) I know how to calculate the wer
but I don"t know how to present it
更新5:
can some1 show me how to present it better?
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-ponent 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 o 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 补充： 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 补充： 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 补充： 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 补充： 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 补充： Because of words limit here...Q41 is given in "opinion (意见) section"
分享好知识！