but when i doing this in lab, my result from lab is quite big difference to the theoretical result. the aim of the lab is to determine the forces in members and the model is same as the diagram. the force applied is also same. What is the possible source of errors?
Do we need to consider the...
Fbc should be tension...i see it...
my understanding about zero force members...
zero force members are a member which cannot carry load. hence there is no force in these members. these members do not have compression force and tension force.
the existence of zero force members are to...
thank you...
Fbc = 20N??
Fbc = Fce cos 45
Fbc = 20N
but why just now what i did is wrong? i feel like is correct because
Fx = 0
Fx contain Cx, Fbc, Fce cos 45, that's why i formed a equation.
Fab = tension
Fae= tension
Fbe = compression
Fde= compression
Fce= tension
Fbc = compression
ok...i got it...thank you
now these are my answer...could you help me check and are they correct?
C joint
(vertical)
Cy - Fce sin45 = 0
20 - Fce sin 45= 0
Fce = 28.28N
(horizontal)
Cx + Fbc - Fce cos 45 = 0
-20 + Fbc - 28.28cos 45 = 0...
i don't understand about roller support there...
how to know the force is in which direction?
i thought the roller support 's force is always oppose to the force applied(20N)...
thank you
[b]1. every member is 0.45m and all the angle is 45degree...
C is fixed support, D is roller support
[b]3. Ʃfx = 0 , Cx = 0
Ʃfy = 0 , Cy + Dy = 20N
ƩMc = 0 , 20(0.45) + Dy(0.45) = 0
Dy = -20N...
[b]1. an 85kg jogger is heading due east at a speed of 3m/s. A 55kg jogger is heading 39degree north of east at a speed of 3m/s. Find the magnitude and direction of the sum of the momenta of the two joggers.
[b]2.i have no idea about this question...can someone tells me what is this...
[b]1.A 0.16kg ball makes an perfectly elastic head on collision with a second ball initially at rest. the second ball moves off with half the original speed of the first ball
[b]2. what fraction of the origianl kinetic energy gets transferred to the second ball? answer=108J
i...
[b]1. A big fish with its mouth wide open is moving at 2.0m/s towards a school of small fish each of mass 300g. the small fish enter the wide mouth of the big one with a velocity of 5m/s to the opposite durection at the rate of 20fish per min. the big one is motionless after 13min : the big one...
what u mean?
but i use Newton's law of restitution to get the answer.
e=(v1-v2)/(u2-u1)
-4.85=v1-v2---------second equation
and i finally get the answer.
[b]1. a 7g bullet is fired into a 1.5kg ballistic pendulum. the bullet emerges from the block with a speed of 200m/s, and the block rises to a maximun height of 12cm. find the initial speed of the bullet and the kinetic energy lost of the bullet.
answer: 528m/s, 835.7j
[b]2. mu+ mu = mv + mv...
[b]1. a 1kg ball traverse a frictionless tube. the ball after falling through a height of 1.2m, strikes a1.5kg ball which initially at rest. find the velocities of the two balls, if the collision is perfectly elastic. V1= -0.97m/s V2= 3.88m/s
[b]2. i found out the initial velocity of...
can i say normal force acting on the object perpendicular to the surface is the frictional force??
because if the normal force is great enough, the object will not slide down...so, can i say frictional force is normal force?
help me...thank you
[b]1
A particle is projected with a velocity of 146m/s at an elevation of 35degree.
max height= 357.8, t=17.1s horizontal distance= 2041.8m
the question is find the velocity and direction of motion at a height of 120m
i got v= 137.72, but what is the meaning of the direction of motion at...
how come straight away get 82.4m...if i use s= vt, s= 23.8cos30 X 4s
= 82.4m
you tell me find the distance from B to C,
1st, i use this formula, s= ut +1/2at^2 to find t
30.8= -11.9t + 4.9t^2
t= 4s and t= -1.57s
2nd , i...
i know is parabolic path.
i don't really understand. what is the x=50? and t= 2.429s? how to get these?
and symmetry means that the middle of the graph,then the middle point is the maximun height, then why the velocity is 23.8m/s ?why not 0m/s?
my solution
1st , i find the the maximun height achieved.
v^2 =u^2 +2as , 0= (23.8m/s sin 30)^2 +2(9.8)s
s= 7.225m
2nd, i find the time taken for achieved maximun height
v=u+at, 0= 23.8 sin 30 + 9.8t
t = 1.214
3rd, i...
[b]1. an object is thrown upwardly from the roof of a building of height 30.8m with an initial velocity 23.8m/s at an angle 30degree to the horizontal. find the horizontal distance from the foot of the building to the point where the object hits the ground. assume g=9.8m/s^2
answer is 82.4m...
ya...thank you...i got it...
i feel quite disappointed to me, because can't solve the question.
any guidances while doing projectile motion's question?
i tried b4
s= vt, s=ut+1/2at^2
t is same
v is same,
so,
s=u(s/v) + 1/2a(s/v)^2
48m= 62m+ 1/2(9.8) (3844/v^2)
v=36.68m/s, the correct answer should be19.8m/s
sorry...i know the equation...too panic...
you mean find the v right now?but i don't know the t.
and i think a= +g, because toward the ground.
the correct answer of this question is 19.8m/s.
but the problem is what is the final velocity? zero or other value??
i know the car falling dowan at 37.75degree from horizontal. i use this formula, tan angle = 48m/62m, hence equal to 37.75degree.
[b]1. a stuntman performs a stunt of speeding off the cliff of a high rise building with the aim reaching another building with a distance away as the distance between two building is 62m and the differences height between two building is 48m. find the minimum velocity that the car must have as...
[b]1. A loaded ore car has a mass of 950 kg and rolls on rails with negligible friction. It starts from rest
and is pulled up a mine shaft by a cable connected to a winch. The shaft is inclined at 30.0°
above the horizontal. The car accelerates uniformly to a speed of 2.20 m/s in 12.0 s and...
okay
1st i find vertical speed. vertical speed = 119.44/tan angle
2nd i find angle, i use this formula, v^2= u^2+2as, 0= (119.44/tan angle)^2 +2(9.8)(1200)
angle= 37.91degree from vertical
and the answer is wrong~
[b]1. a rescue plane is flying at a constant elevation of 1.2km with a speed of 430km/h toward a point directly over a person struggling in the water. at what angle of sight should the pilot release a rescue capsule if it is to strike the person in the water...answer is 57.3degreen from...