Calculate the forces FA and FB that the supports exert on the diving board of Fig. 9–42 when a 58-kg person stands at its tip. (a) Ignore the weight of the board. (b) Take into account the board’s mass of 35 kg. Assume the board’s CG is at its center.
Step 1 of 3 The impulse is large force acts on a body in short interval of time We know that p F t p F =k t where kis proportional constant its value is 1 F = mv because linear momentum p=mv t mv F = t here m is constant F t=mv F t=p where p=p -f i J = p -p f i J =mv- mu This theorem is called as impulse momentum theorem Step 2 of 3 Given u=5.5 m/s v=10.4 m/s J = p=p -p f i =mv- mu =m(10.4- 5.5)m/s = m 4.9m/s If F t=p Then F avg= p t m(4.9m/s) = t From second law F avg a= m m(4.9m/s) t = m t= 4.9 m/s a t= 4.9 m2s 9.8 m/s t=0.5 sec The stone will take 0.5 sec to increase its speed from 5.5 to 10.4 Step 2 of 3 Step 3 of 3