00:01
In this problem, we'd first like to find a free body diagram of our beam.
00:06
So here we have our beam, and at our right -hand side, we have a tension force, or left -hand side.
00:12
On our right -hand side, or in the middle, we have our weight force, and breaks into f -y and f -x.
00:21
Now, on our right -hand side, we have our different force is f -y and f -x, so f -x is 5 ,000 times cosine of 4 -g.
00:31
Degrees and the y component is the sign we're also given e is 200 times 10 to the 9 noon per meter squared and our mass is 50 kilograms then we like to calculate the tension of steel cable oa so let's first balance our torque about c our torque times 5 is equal to mg times 1 plus 5 000 sine of 40 times 3.
01:06
So we have 5t is equal to 50 times 9 .8 plus 5 ,000 times sine of 40 times 3.
01:17
So our torque is 2026 .0 .3 newtons.
01:24
For part c, we'd like to determine the reaction force.
01:29
Now for our reaction force, we have t plus f y, plus 5 ,000 sine of 40 is equal to m times g...