00:01
So we have here a rod that is 2 meters long.
00:14
At the center of the rod is where its weight pulling it downwards.
00:23
So the weight is given by with a value of 255 newton's weight of the rod.
00:40
On the right end of the rod, a weight is hanging with a magnitude of 225 newtons.
00:54
A weight w .5 newton from the left end is also hanging.
01:03
Which is denoted only as w.
01:07
And the fulcrum of the rad is located 0 .75 centimeters, 5 .75 meters from the right end.
01:23
So this is the full chrome.
01:25
Then we can say this is 0 .25 meters, and this is also 0 .5 meters.
01:35
Now, question a, asked for the magnitude of the weight on the left end which is weight w, wherein the rad is in balance position.
01:52
The rod is in balance if the torques on the left side of the rod is equal to the torque on the right side of the rod.
02:04
So for the left side, we have the weight multiplied by a distance from the fulcrum, which is 0 .5 plus 0 .25, which is 0 .75 meter plus the weight of the rad which is 255 newtons multiplied to its distance from the fulcrum which is 0 .25 meters they are equal to the torque on the right side which is only the hanging at the right end which is 225 newtons and its distance from the full crew is 0 .0 .0 is 0 .0 .0 .0 is 0 .0 .0 is 0 .0 is 0.
02:51
Point 75 meters.
02:53
So we're arranging and isolating w to the left side equation since this we're looking for then we have w is equal to this.
03:17
So calculating we can find that w is equal to 140 newtons.
03:29
Now for situation b what if this weight is move 20 .25 meters to the right...