00:02
In this problem, you have a boom that's holding up a weight on its end.
00:07
It's supported by a hinge at what i call point a and a cable.
00:13
And the goal of the problem is to find the tension in the cable and the force at the hinge, which is point a here.
00:20
And the way we do this is we start with a free biore diagram for the boom.
00:25
Now we have the weight.
00:28
We have the weight on the end.
00:29
So there's a tension in the cord, but it's equal to the weight of the, object in the end, so i'll just call it w here.
00:37
I'll just be w.
00:38
Now the boom itself is a uniform object that means its weight is acting in its center.
00:47
So this is wp.
00:49
We have the tension, always pulling, and we are told that it makes a 45 degree angle with the horizontal, so this is the tension.
01:00
And we have the f -a, this is point a, f -a -y, and we have f -a -x.
01:12
They could be in the opposite direction, they could be zero, that we'd find out from the equations.
01:19
So that's our free by diagram that covers everything.
01:23
Also notice, boom makes a 20 -degree angle with the horizontal.
01:27
So you have all that.
01:28
So, this is an equilibrium problem.
01:31
F -net -x equals zero, f -n -n -y, equals zero, net torque equals zero.
01:36
So we'll take each one in turn.
01:38
F -n -x, f -a -x, minus t -cosine -45 degrees, is equal to zero.
01:47
It's equation one.
01:48
Can't do anything with it because of the two unknowns.
01:53
F -n -y, f -a -y, minus, or actually, let me put to the tension one first.
02:01
T -sign -45, minus w -b, minus w -b, minus w -w, is equal to zero, it's equation two.
02:11
Again, two unknowns.
02:14
But once we have tension, everything falls into place easily.
02:19
Now, the net torque equation.
02:22
Now, that's what we have this little table over here.
02:24
Now, the torque is plus or minus the force times its moment arm.
02:28
Plus, if that force return to counterclockwise, minus if the force return it clockwise.
02:35
Now, remember the moment arm.
02:37
Can view it two ways.
02:37
The perpendicular distance between the line of action of the force and the rotation axis, or another way thinking about the shortest distance from the rotation axis to the line of action.
02:49
Either way.
02:50
So let's look at each one.
02:51
Remember, the line of action is an infinite line that goes through the force.
02:56
So there is a line of action for f -a -y and f -a -x.
03:00
Here's the line of action for wb.
03:03
There's a line of action for the tension, and here's the line of action for the weight.
03:07
Let's look at f -a -x, f -a -y.
03:11
Notice, let's use the shortest distance version.
03:17
You're standing at the rotation axis.
03:20
And what's the shortest distance to get to the line of action of f -a -x? you look down, you're standing on it.
03:26
Zero.
03:28
Likewise for f -a -y, zero.
03:30
So there is no rotational effect from either one of those.
03:34
So they're out.
03:34
That's why i chose point a as my rotation axis.
03:37
I chose another point, then they appear.
03:40
And i got very heavily coupled equations.
03:43
Not more mathematical work to do.
03:45
Why do extra work when you don't have to? that the physics help you...