nice clear explanation. I'm reviewing the subject, but enjoyed this. Looking forward to the entire play list and then checking out some of your other content. Cheers!
where are u getting these answers from I've tried multiple times and after inputting the same info you did i got -33 on the first and -11 on the second
thenitap for torque start with these:czcams.com/video/zoDGTamogxM/video.html czcams.com/video/r2OYfK8ukdY/video.html czcams.com/video/DPTC1Txa9Wo/video.html czcams.com/video/f_u2dPs-UM8/video.html Then you can try a few on static equilibrium czcams.com/video/R57_Ax9MD1w/video.html czcams.com/video/dL69oscyb8A/video.html
Imagine that the force vector is the hypotenuse of a right triangle; you can split this vector into its horizontal and vertical components, representing the legs of your triangle. You know theta, which is the angle between the radius of the lever arm and the force. Sine of theta is the ratio of the length of the leg opposite the angle over the length of the hypotenuse, or in this case, the magnitude of the perpendicular force to the magnitude of the actual force. Basically, sine of theta will always produce the component of the force perpendicular to the lever arm while cosine will produce the component that is parallel. Hope that wasn't confusing or condescending. :)
Nicholas O'Brien The right triangle idea really breaks down for me when we are working with angles greater than 90. Right now to understand this i break up the angles until i have a right triangle that makes sense to me and then work with that. Using sine woth angles greater than 90 while thinking about right triangles gives me a bit of a headache. For example right now im working with a 4.3m lever beam with a 65N force at the end of the beam that is pointed 150 degrees above the perpendicular. The angle is written on the side closer to the pivot point so that the force is pointing out if that makes sense. If the angle was on the other side and read 30 degrees it would make more sense to me. I would use sin of 30 to figure out the ratio between the hypotenuse 65N and the vertical component of the force. Can I go straight from the 150 degree angle to the vertical component? Edit: I can see that sine of 150 and sine of 30 are both 1/2 but it still doesnt sit right with me. Will that relationship between theta and 180-theta always work?
I Had No Idea What Torque Is Since I Missed My Physics Lecture....Thank You So Much And You're Way Better Than My Lecture :)
molo Naledi
nice clear explanation. I'm reviewing the subject, but enjoyed this. Looking forward to the entire play list and then checking out some of your other content. Cheers!
Thanks for watching and commenting.
You sir are an amazing teacher- Thank you very much-
Thank you for the positive comment.
You can link to all my videos at my website: www.stepbystepscience.com
amazing lecture! Thank you very much for it.
That's all what it takes folks.
5 min of this guy!
I always try to keep it short and get right to the point.
You can link to all my videos at my website: www.stepbystepscience.com
Thank you!
This makes much more sense now. Thank you!
Thank you! Very helpful!
This Guy Is Too Legit Too Quit!!! :) ❤️❤️😆😆
Fuzz Fuzz Thank you for the nice comment, not planning on quitting yet!
where are u getting these answers from I've tried multiple times and after inputting the same info you did i got -33 on the first and -11 on the second
I believe you have your calculator set on radians, it should be set to degrees. Try it and let me know.
Thank you for the information 👍🏼👍🏼👍🏼
You are very welcome.
Thank you so much
Thank you.
Thank You
F is the Force. Indeed!
very good thank you
+Jade Gretton You are welcome.
You can link to all my videos at my website: www.stepbystepscience.com
i dnt get why u use 130º angle in that 1st example... why not 50º?
You can use either one. I just try to be consistent in how I apply the rule for the angle between the lever arm and the force.
thank you Sir, I get it.
What is the beginning video to build on I have to start from # 1 beginner
thenitap for torque start with these:czcams.com/video/zoDGTamogxM/video.html
czcams.com/video/r2OYfK8ukdY/video.html
czcams.com/video/DPTC1Txa9Wo/video.html
czcams.com/video/f_u2dPs-UM8/video.html
Then you can try a few on static equilibrium
czcams.com/video/R57_Ax9MD1w/video.html
czcams.com/video/dL69oscyb8A/video.html
sir I've some confusion about theta . that's why we use sine thta why not cosine theta
Imagine that the force vector is the hypotenuse of a right triangle; you can split this vector into its horizontal and vertical components, representing the legs of your triangle. You know theta, which is the angle between the radius of the lever arm and the force. Sine of theta is the ratio of the length of the leg opposite the angle over the length of the hypotenuse, or in this case, the magnitude of the perpendicular force to the magnitude of the actual force. Basically, sine of theta will always produce the component of the force perpendicular to the lever arm while cosine will produce the component that is parallel. Hope that wasn't confusing or condescending. :)
Nicholas O'Brien
The right triangle idea really breaks down for me when we are working with angles greater than 90. Right now to understand this i break up the angles until i have a right triangle that makes sense to me and then work with that. Using sine woth angles greater than 90 while thinking about right triangles gives me a bit of a headache.
For example right now im working with a 4.3m lever beam with a 65N force at the end of the beam that is pointed 150 degrees above the perpendicular. The angle is written on the side closer to the pivot point so that the force is pointing out if that makes sense. If the angle was on the other side and read 30 degrees it would make more sense to me. I would use sin of 30 to figure out the ratio between the hypotenuse 65N and the vertical component of the force. Can I go straight from the 150 degree angle to the vertical component?
Edit: I can see that sine of 150 and sine of 30 are both 1/2 but it still doesnt sit right with me. Will that relationship between theta and 180-theta always work?
i input that information on a calculator and it gave me -33
me too
You also probably have you calculator set on radian, it should be on degrees.
why so much blablablabla. so much talking. an image shows it all.
50/50