Electric forces and Electric fields question on x and y components

AI Thread Summary
The discussion centers on confusion regarding the classification of components in a physics problem involving electric forces and fields. The original poster questions why T cos theta and mg are labeled as x components when they appear to be on the y-axis. Participants clarify that the designation of components depends on the orientation of the coordinate system used in the problem. Misunderstandings often arise from rigid interpretations of trigonometric functions, where cosine is associated with horizontal (x) and sine with vertical (y) components. Emphasis is placed on understanding the geometry of the problem and the correct application of trigonometric definitions to resolve such confusion.
mattmro
Messages
2
Reaction score
0

Homework Statement



please see the pictures!

Homework Equations



F=k(q1)(q2)/r^2



The Attempt at a Solution



okay, I am just curious as to why t cos theta and mg are x components when they are clearly on the y axix!
I thought you used cos for x components and sin for y components
 

Attachments

  • dd.jpg
    dd.jpg
    33.9 KB · Views: 512
  • ddd.jpg
    ddd.jpg
    35 KB · Views: 465
Physics news on Phys.org
guys I know I am a DA, i barely got by in physics I and I was able to get by without knowing how to do these type of problems, now I can't please help!
 
The cosine of theta is the adjacent side of the triangle divided by the hypotenuse. With theta defined to be where it is in this problem, T cos theta is the vertical component of the tension.
 
mattmro said:
okay, I am just curious as to why t cos theta and mg are x components when they are clearly on the y axix!
I thought you used cos for x components and sin for y components
Where is it written that they are on X axis ... looks on Y to me
Maybe they took X as Y and Y as X

mikelepore said:
The cosine of theta is the adjacent side of the triangle divided by the hypotenuse. With theta defined to be where it is in this problem, T cos theta is the vertical component of the tension.
OP's asking why the vertical component is called X component
read question again
 
cupid.callin said:
OP's asking why the vertical component is called X component
read question again

I already read the question and the images about ten times before answering. It sounds to me like a teacher may have pounded it into the student's head that x means horizontal and then always take the cos, y means vertical and then always take the sin. I have heard some teachers say that, although it is wrong. The cure for the confusion is to identify the triangle and remember the "sohcahtoa" definitions of sin and cos.
 
TL;DR Summary: I came across this question from a Sri Lankan A-level textbook. Question - An ice cube with a length of 10 cm is immersed in water at 0 °C. An observer observes the ice cube from the water, and it seems to be 7.75 cm long. If the refractive index of water is 4/3, find the height of the ice cube immersed in the water. I could not understand how the apparent height of the ice cube in the water depends on the height of the ice cube immersed in the water. Does anyone have an...
Thread 'Variable mass system : water sprayed into a moving container'
Starting with the mass considerations #m(t)# is mass of water #M_{c}# mass of container and #M(t)# mass of total system $$M(t) = M_{C} + m(t)$$ $$\Rightarrow \frac{dM(t)}{dt} = \frac{dm(t)}{dt}$$ $$P_i = Mv + u \, dm$$ $$P_f = (M + dm)(v + dv)$$ $$\Delta P = M \, dv + (v - u) \, dm$$ $$F = \frac{dP}{dt} = M \frac{dv}{dt} + (v - u) \frac{dm}{dt}$$ $$F = u \frac{dm}{dt} = \rho A u^2$$ from conservation of momentum , the cannon recoils with the same force which it applies. $$\quad \frac{dm}{dt}...
Back
Top