Understanding when to use sine and cosine in force problems can be simplified by visualizing the situation with a right-angled triangle. The mnemonic SOH CAH TOA helps recall that sine relates to the opposite side and cosine to the adjacent side of the angle. If the angle is measured from the horizontal, sine is used for the vertical component, while cosine is used for the horizontal component. Conversely, if the angle is from the vertical, cosine applies to the vertical component and sine to the horizontal. Utilizing these principles can significantly aid in solving force vector problems effectively.
#1
AnthroMecha
26
0
I am having the hardest time attaching my brain to some sort of method to know when to use sine and cosine on force problems. What is an easy way of remembering which function to use to find the force in the direction of x and force in the direction of y?
I am having the hardest time attaching my brain to some sort of method to know when to use sine and cosine on force problems. What is an easy way of remembering which function to use to find the force in the direction of x and force in the direction of y?
First of all, draw a picture of the right-angled triangle, so that you don't have to just visualize it in your head.
Second of all, use the mnemonic SOH CAH TOA to remember the definitions of the trigonometric ratios.
Sine = Opposite/Hypotenuse.
Cosine = Adjacent/Hypotenuse.
Tangent = Opposite/Adjacent.
Third of all, realize that in decomposing force vectors, there are usually only ever TWO possible cases.
Case 1: The angle that you've been given is measured from the horizontal
In this situation, Fy is the side of the triangle that is opposite from the angle, and Fx is the side of the triangle that is adjacent to the angle. (The total magnitude, F, of the force, is always the hypotenuse). Therefore, it follows that:
sinθ = Fy/F (opposite side / hypotenuse)
cosθ = Fx/F (adjacent side / hypotenuse)
Fy = Fsinθ
Fx = Fcosθ
Case 2: The angle that you've been given is measured from the vertical
In this situation, Fy is the side of the triangle that is adjacent to the angle, and Fx is the side of the triangle that is opposite from the angle. (The total magnitude, F, of the force, is always the hypotenuse). Therefore, it follows that:
sinθ = Fx/F (opposite side / hypotenuse)
cosθ = Fy/F (adjacent side / hypotenuse)
Fy = Fcosθ
Fx = Fsinθ
So, you can see that, if the angle is measured from the horizontal, then the cosine is associated with the horizontal component, and the sine is associated with the vertical component.
if the angle is measured from the vertical, then the cosine is associated with the vertical component, and the sine is associated with the horizontal component.
What is an easy way of remembering which function to use to find the force in the direction of x and force in the direction of y?
I keep telling people …
it's always cos!
It's alwayscos of the angle between the force and the direction …
whenever it looks like sine, that's because you're using the "wrong" angle …
maybe θ is marked on the diagram, but if the angle you really want is 90°-θ, then you use cos(90°-θ), which of course is sinθ !
(however, a good check, when you're using slopes, is to imagine "what would happen if the slope was 0°?" … would the component vanish (sin0°) or be a maximum (cos0°) ?)
#4
AnthroMecha
26
0
This forum always delivers. Thanks guys these are very useful tools.
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}...
I was thinking using 2 purple mattress samples, and taping them together, I do want other ideas though, the main guidelines are;
Must have a volume LESS than 1600 cubic centimeters, and CAN'T exceed 25 cm in ANY direction.
Must be LESS than 1 kg.
NO parachutes.
NO glue or Tape can touch the egg.
MUST be able to take egg out in less than 1 minute.
Grade A large eggs will be used.
, when you're using slopes, is to imagine 
"what would happen if the slope was 0°?" … would the component vanish (sin0°) or be a maximum (cos0°) ?)
