Three point charges are located at the corners of an equilateral triangle as in Figure P15.13. Find the magnitude and direction of the net electric force on the 2.00 µC charge.
Love the simplification of this. Unlike most it seems, I've got a great physics prof. But dumbing things down even further like this makes things a lot easier.
in some problems when finding the component you switch what trig function you use to find the components in this one you use sin for the x component why wouldn't it be Cos doesn't cos usually correspond with x?
cos doesn't correspond with x, it corresponds to the adjacent over hypotenuse. The adjacent side to your angle could be the y, it doesn't have to be x. It just depends on where the angle you are using is
i think the procedure to do this is helpful and correct .. but the answers are wrong answers... why Fx has to be 3.6 yet when you calculate what is visible on the video you get 5.36 ?
Three charges A(4 µC), B(-6 µC) and C (2 µC) are placed at the vertices of a right angle triangle ABC. AC = 10 cm, BC = 6 cm. Find the net force on charge B due to Charge C and A.
cos doesn't correspond with x, it corresponds to the adjacent over hypotenuse. The adjacent side to your angle could be the y, it doesn't have to be x. It just depends on where the angle you are using is
Great video, it would be also nice though to see the actual numbers at 5.06. I do not know why but for some reason I keep getting the wrong answer for the sum of the x forces even though my calculator is in degree mode.
Is it necessary to put cosϴ and sinϴ while calculating ∑fx and ∑fy or can we calculate them without involving Sin and Cos in this by the taking the product of Coulomb's constant and charges?? Please tell me!
For some reason my brain thought I should find the electric field and I spent an extra 30 minutes trying to figure out what I did wrong before I read the question again
Can you please tell me how you got the total of sum of forces for the x direction, I plugged the #s on my calculator but I'm not getting 3.6x10^2!!!! :/
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Yes but that is in the conventional scenario where the y-axis is opposite the angle represented in the diagram. In this case however, the y direction (pointing downward) is adjacent to the angle 30°.
cos(30) is equal to sin(60) and vise versa. Therefore you can either use cos(30) to get your value for the adjacent side or sin(60) for the opposite side and get the same value.
2 years late but here you go- Usually the way it is done is that x vectors will use cos(theta), and y components sin(theta) because they use the angle the vector makes with the x-axis. He used the angle made with the y axis instead for some reason, so the trig is flipped. As you can see, sin30 = cos 60. Cos60 is what i recommend doing.
A turning folk vibrating at 512Hz falls from rest and accelerates at 9.80m/s^2. How far below the point of release is the turning fork when waves if frequency 485Hz reach the point of release?
Because he calculated them using the opposite angle (30). you could use the 60 degree angle and have sin(60) in your equation to get the same value as cos(30), just depends on your preference.
4. A cylindrical water tower of diameter 3.0 m supplies water to a house. The level of water in the water tower is 35 m above the point where the water enters the house through a pipe that has an inside diameter 5.1 cm. The intake pipe delivers water at a maximum rate of 2.0×10−3m3⋅s−1. The pipe is connected to a narrower pipe leading to the second floor that has an inside diameter 2.5 cm . What is the pressure and speed of the water in the narrower pipe at a point that is a height 5.0 m above the level where the pipe enters the house? please do this I have an exam
The way you resolve components is wrong for X axis we use cos not sin and for y components we use sine not cos...that's why people gets wrong solutions
