IMPACT UPON BEAMS. 
99 
Problem 2. To find the deflection of the beam answering to 
a given horizontal impact, and the time of its duration. 
Put c J = the chord of the arc through which the ball fell, 
v' = the velocity of impact, h =■ the height fallen through, and 
the rest as before. 
Since from the last problem, b y g ^ ^ = the velocity 
which the beam would generate, in the ball and itself, while re¬ 
covering its form, this quantity will represent the velocity de¬ 
stroyed in both in bending the beam through the same distance b . 
Multiplying, therefore, the above by w -f- r gives 
b (tv + r) A / / h ^ —- = the momentum destroyed. 
v ’ V e (w -f r) 
But v'tv = the momentum of the impact. And since these are 
equal, 
whence 
V 1 W = b (tV -f ?') A / ; -r 5 
v ’ V e [tv + r) 
= v' w 
£P (w + r) 
But = a/2 g h, from the properties of falling bodies. 
, / 2 
b — w A / —t- 
V p hi 
h e 
j) (■tv -f r ) 
And since h = 
c' 2 
TV 
tv c 1 
sZ \ 
p l {tv + r) % 
From the three last equations we have the deflection in terms 
of the velocity of the impact, the height, and the chord of the 
arc fallen through, the beam and ball being the same. 
For the time, or the duration of an impact.—The time of de¬ 
flecting the beam, through any distance b , must be equal to the 
time of returning through the same distance ; the beam being 
supposed perfectly elastic. And since the time of return was 
/ C (iV "I” ? j 
found, by the last problem, to be 1'57079 x -y/ — 
/g h(j 
therefore twice this quantity or 3T4159 x /y/ -—-= the 
time of an impact, or complete vibration of the beam with the 
striking body accompanying it. The time is therefore constant 
