00:01
For this problem, we're going to determine the reactions at point a and b for three different alpha values.
00:07
Part a, alpha equals zero degrees.
00:09
Our first principle, we know that because we are in static equilibrium, the sum of the torques at point a is equal to zero.
00:17
We can substitute this in for b from 0 .5 meters minus the 100 -newton meter force is equal to zero.
00:28
Solving us for b gives us 200 newtons.
00:37
We also know the forces in the x direction must sum to zero.
00:41
We can simply substitute this in for a sub x is equal to zero because there are no forces present there.
00:50
The sum of the forces in the y direction are also equal to zero.
00:53
This we substitute a sub y plus 200 is equal to zero.
01:01
Solving gives us a equal to 200 newtons.
01:07
Downwards.
01:11
Part b, we see our alpha value is now 90 degrees.
01:15
Our rotational torque must sum to zero once more.
01:19
In this instance, we can substitute in b at 0 .3 meters minus the 100 -newton meter force is equal to zero...