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  1. S

    FRAME work ( statics)

    thank you very much...you help me a lot...appreciate your help...(bow)
  2. S

    FRAME work ( statics)

    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...
  3. S

    FRAME work ( statics)

    Fbc = 0N?? is not tension and compression... then why this phenomenon happenned? i mean zero force members...
  4. S

    FRAME work ( statics)

    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...
  5. S

    FRAME work ( statics)

    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
  6. S

    FRAME work ( statics)

    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...
  7. S

    FRAME work ( statics)

    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
  8. S

    FRAME work ( statics)

    [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...
  9. S

    Momentum and collision

    [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...
  10. S

    Momentum and collision

    [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...
  11. S

    Momentum and collision

    [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...
  12. S

    Momentum and collision

    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.
  13. S

    Momentum and collision

    https://www.physicsforums.com/showthread.php?t=519410 can you help me this question?
  14. S

    Momentum and collision

    i got it...thank you very much~
  15. S

    Momentum and collision

    [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...
  16. S

    Momentum and collision

    [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...
  17. S

    Friction on inclined plane

    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
  18. S

    Projectile motion

    [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...
  19. S

    Projectile motion of a thrown object

    okay...thank you very much...
  20. S

    Projectile motion of a thrown object

    why gravity should be -9.8m/s^2, the object is toward gravity , right?
  21. S

    Projectile motion of a thrown object

    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...
  22. S

    Projectile motion of a thrown object

    the answer become 25m again. t= 1.214 from quadratic equation, s = 25m ..
  23. S

    Projectile motion of a thrown object

    i understand 1st and 2nd paragraph, but in 3rd paragraph, -23.8sin 30 means what? and what you want me to do? find what?
  24. S

    Projectile motion of a thrown object

    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?
  25. S

    Projectile motion of a thrown object

    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...
  26. S

    Projectile motion of a thrown object

    [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...
  27. S

    Projectile motion of a stuntman

    actually i quite confuse with this, sin and cos?? when i should use sin angle and cos angle??
  28. S

    Projectile motion of a stuntman

    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?
  29. S

    Projectile motion of a stuntman

    hah? then t = 62s? then s= 1/2at^2, where is the v?
  30. S

    Projectile motion of a stuntman

    but i don't know what is v on horizontal line...s=vt, v=?
  31. S

    Projectile motion of a stuntman

    then you mean 48= 0 +1/2(9.8)(62/v cos 37.75)^2 if like this, i also can't get the answer
  32. S

    Projectile motion of a stuntman

    ya...s =vt...horizontal moving in the constant velocity. this one i got it.
  33. S

    Projectile motion of a stuntman

    having dizzy... still can't figure it out...
  34. S

    Projectile motion of a stuntman

    sorry, i do not understand...is the horizontal initial speed is not same as vertical initial speed?
  35. S

    Projectile motion of a stuntman

    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
  36. S

    Projectile motion of a stuntman

    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.
  37. S

    Projectile motion of a stuntman

    you mean should be parabolic path?? ya... i know... v^2=u^2+2as v=u +at s=ut+at^2
  38. S

    Projectile motion of a stuntman

    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.
  39. S

    Projectile motion of a stuntman

    [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...
  40. S

    What maximum power must the motor provide?

    pulling force...power = workdone / time, workdone = force x distance
  41. S

    What maximum power must the motor provide?

    i am finding this answer (a) What power must the winch motor provide when the car is moving at constant speed?
  42. S

    Projectile motion

    thank you, i got it... thank you very much...
  43. S

    Projectile motion

    how to find the vertical velocity?
  44. S

    What maximum power must the motor provide?

    [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...
  45. S

    Projectile motion

    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~
  46. S

    Projectile motion

    no~i mean i should take vertical speed or horizontal speed?
  47. S

    Projectile motion

    what is the speed i should take?
  48. S

    Projectile motion

    [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...
  49. S

    Projectile motion

    sorry...mistake again...i know how to do already, just now made some mistakes in calculation. thank you very much...
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