00:01
Here's an example where we're going to need both translational and rotational equilibrium ideas.
00:09
Both of these involve newton's second law, but the translational equilibrium states that we must have the sum of forces in the x direction equals zero balancing.
00:27
The sum of the forces in the y direction or the y components must also balance.
00:35
And for rotational, we must have the sum of torques equals zero.
00:42
And a reminder that a torque is a product of a direction counterclockwise or clockwise, and then a distance out from the pivot times the force, times the sign of the angle in between the two, and it has either plus or minus, depending on which way the object would tend to rotate clockwise or counterclockwise in a plane at least.
01:13
So to work with this, we will need to have a force diagram on this particular beam.
01:19
It is making an angle of 30 degrees with respect to the horizontal.
01:25
The rope is making a 40 degree angle to the beam.
01:32
But in terms of our forces, we have the tension in the rope pulling at six meters away from the wall.
01:43
So there is a wall there.
01:45
Let's go ahead and show the wall.
01:47
There is the weight that plays.
01:52
Pulls straight down.
01:55
We're assuming a uniform beam.
01:58
So we have mg at a distance of four meters.
02:04
So yeah, we've got some different r's coming into play.
02:08
And finally, at the wall itself, there is going to be a normal force out from the wall that will push straight out.
02:19
In addition, there is some friction, static friction.
02:25
I'm going to show that pointing up.
02:27
But our formula will tell us whether that's working or not.
02:34
In order to do torques, we will have to pick a pivot.
02:38
And usually what i like to do is pick the spot that has the most unknowns on it.
02:46
And so i will choose my pivot for calculating torques to be at the wall.
02:54
So our tension is the force that we'll have to decompose into two components, x and y.
03:04
The y component is t, cosine, sorry, sine theta is 10 degrees, sine of 10 degrees.
03:14
And our x component is tension cosine 10 degrees.
03:24
Okay, so we have three equations, and hopefully we can solve for all the three unknowns that we have in the problem.
03:35
So the first equation i'll write down is the sum of the x components, and i usually like to balance them right to left.
03:45
So i see fn is pushing to the right...