T2Cos45+10N = 0 turning out to be negative is not the same thing as a directional indicator. The fact that 10N in the Fx component should be subtracted is something else. T2Cos45+10N is solving for the tension of T2 not setting the direction of 10N in Fx. The point is that if 10N is really...
A ten Newton block is hanging from a line, goes up and over a pulley and to the right at a 90 degree angle. It hits an equilibrium point. At this eq point a line leaves upwards of tension T2 at 45degrees. Also at this eq point a string hangs down of unknown N.
Here are the components to...
I see that a(sqr) + b(sqr) = c(sqr) is for finding the tension of T2. This shouldn't work if the angle is anything but 45 degrees. If we altered the weights it still shouldn't make a difference if the line angles are making a right triangle. I thought the pythagorean thereom was for finding...
There is an error in my last message. The statement should be a(sqr) + b(sqr) = c(sqr). This alters the math to the following:
10(sqr) + b(sqr) = c(sqr)
100 = c(sqr) - b(sqr)
10 = c - b
Again, the answer is ten, but i have an extra variable.
"Cross my heart" -M 1988
Backtracking: I have a question. The first given weight at the very beginning of the problem is hanging down and has a weight of 10N. It goes up and over a pulley and makes a right angle and goes straight to the right. At this point it hits an equilibrium point where a block is hanging...
Could T2 in Fx be negative because it is considered to be in the trigonometric quadrant 2 making cosine a negative if the angle is greater than 90 degrees? You would have to consider T3 in quadrant 1 and the weight on the Y axis. Cosine and sine do not come into play for the weight since it...
For the following components i have 2 questions:
Fx = -T2cos(30)+T3cos(50) = 0
Fy = T2sin(30)+T3sin(50)-N = 0
First, in Fx, T2 is negative. In Fy, T2 is positive. Further, in Fx and Fy both T3 are positive. Is this because the y components are parallel lines to one another and are not...
If i know that T2 is 14.1N do i need to create x and y components for it such as follows:
T2x = Cos30 (20N)
T2y = Sin30 (20N)
The magnitude must be 20N since this is what is coming from that side. This seems redundant and is obviously incorrect. We know the tension of T2 is 14.1, it is not...
I revised my message with a better description of the problem in edit. If somebody could please check it over again. I think you can get the components from it more easily. It is message #7. Thanks.
Yes, you got the whole thing correct in your description except you missed one thing. Let
me explain the diagram fully:
(Note the names of the strings as the book is different then what we have labeled them as)
A 10N block is hanging from T1 all the way out at the left. It goes up and over a...
Thanks. Let me clarrify. The string going off to the left from the equilibrium point goes over a pully and comes back down. There is hits another equilibrium point. At this point a weight of 10N is hanging straight down. (The angle up to the pully is 45 degrees on this new side). From this...
I am assuming that the following components are correct:
0 = Rx = 14.1N - T3Cos 50 - ?N
0 = Ry = 14.1N - T3Sin 50 - 20N
If i do what you said and solve for ?N in Rx i will still have the unknown T3 and the unknown ?N. The book problem was easier because there was no y component so i believe T3...
I have an example and a problem i am working on. The example is as follows: There are three strings. One hangs down with 100N while of the other two one goes up to the right from the equilibrium point at 45 degrees and the other goes left. I believe this set of Rx and Ry vector components is...