Because we took the downward direction to be positive, so the weight force time will be considered positive because its direction is also downwards. However, in the first diagram, we took upwards to be positive and weight was downwards therefore we considered it negative. I hope this helps.
in a very simple version of this, 10kg on either side, stationary, the tension in the cable is near equal to 10kg/98N? not 20kg/196N? how is this possible? does no point in the cable experience 20kg?196N
Pretty sure in this situation the pin connecting the cable to the wall would experience 196N worth of force, while the cable tension equals 98N. U could prove this by drawing an FBD about the two masses and then one about the pulley itself. While the pulley experiences 2 downward forces of C, the masses themselves each experience 1 upward force of C.
![กี่หมื่นฟ้า | Your Sky Series EP.1 [2/4]](http://i.ytimg.com/vi/qJbQs_nhjD4/mqdefault.jpg)
You are my savior. I'm going to ace my test tomorrow thanks to this.
I hope it went well! 🙌
Because of you I got full score on my physics retake 🙌🙌
thank you so much, my teacher didn't explain that well so I didn't understand, but you explained it very simply!
If only you were my lecturer
Me too. At least I can help you here though!! 🤜🤛
Thank you 😊
pro saved my life
Thank you
Youre amazing
Thanks Shana 😊
Why is weight (w2) on m2 positive???
imagine the string is strung out in a straight line. The acceleration will be pointing in the same direction
to make the calculations easer you change the m2 direction to + down
Because we took the downward direction to be positive, so the weight force time will be considered positive because its direction is also downwards. However, in the first diagram, we took upwards to be positive and weight was downwards therefore we considered it negative. I hope this helps.
Why is the tension the same for both masses? Is the tension always the same in these problems?
because its the same string
I'm very confused by what's going on with the signs of the numbers
Parker Union
Reinhold Hill
i don't get the 28kg part, where did that come from?
he just grouped 11kg a and 17kg a and added them together
in a very simple version of this, 10kg on either side, stationary, the tension in the cable is near equal to 10kg/98N? not 20kg/196N? how is this possible? does no point in the cable experience 20kg?196N
Pretty sure in this situation the pin connecting the cable to the wall would experience 196N worth of force, while the cable tension equals 98N. U could prove this by drawing an FBD about the two masses and then one about the pulley itself. While the pulley experiences 2 downward forces of C, the masses themselves each experience 1 upward force of C.
Benjamin Valleys
same as a= (m2-m1/m1+m2)g