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it,s telling the downward force ...there cant be upward force /tension due to load and force of motor as if there were force what thing wud prevent the spring from being pulled in tht direction.In case of downward force(weight) on the spring it is counteracted by the load(on the other side) -
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Ur answer to Q11 seems correct but how,d u arrive at answer B in question 12.For me it,s as simple as C ..They asked for instantaneous power which can be observed thru the gradient and tht is most steep from t=2 to 3 -
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I think it,s B as 100 -20 =80 N downwards on right hand side..So to pull at 50 rev/s in opposite dir which means at 15m/s(circumference times rev) wud give power as Force *velocity i.e 80*15 =1.2kW -
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Acceleration is change in velocity per unit time.. Case 1 :If velocity is decreasing then acceleration is at opposite side of the velocity..Case 2:When turning a corner(circulr arc) at constant speed .Instantaneous velocity is tangent to circular arc and acceleration is towards the center e.t.c -
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Well u do know tht option C and D r out of the way as the value of a wud decrease over time.I think this question refers to the happenings on the latch at an instant.For example we know tht at the release pt,where spring is most compressed,instantaneous force is most and as the spring... -
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Da Hell is with u and FLRNAB liking every post . And yeah thnx for tht by the way :P -
Nah it,s pretty easy,U can do it urself even :P
Nah it,s pretty easy,U can do it urself even :P- M.Omar
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Glad to know u r done hibernating :D
Glad to know u r done hibernating :D- M.Omar
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why don't u come tomorrow and see it for urself :D
why don't u come tomorrow and see it for urself :D- M.Omar
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ratio: -7/5 or sth like tht..... right?
ratio: -7/5 or sth like tht..... right?- M.Omar
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haha .....but seriously how hard was tht question..u r just spurring me all the more to see tht...
haha .....but seriously how hard was tht question..u r just spurring me all the more to see tht question now :D- M.Omar
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