+ or - in an equilibrium problem

[blutoonwcarrotnnail]

A ten Newton block is hanging from a line, goes up and over a pulley and to the right at a 90 degree angle. It hits an equilibrium point. At this eq point a line leaves upwards of tension T2 at 45degrees. Also at this eq point a string hangs down of unknown N.

Here are the components to find the second block:

Fx = T2Cos45 + 10N
Fy = T2Sin45 - N

This yields a T2 of 14.1 which is plugged into Fy yielding an N of 9.96.

Why is the 10N in Fx added and not subtracted? The component for it (T1x) is 180 degrees from its origin of the x-axis in quadrant 1. Cosines in quadrant 2 and 3 in trigonometry are both negative. Why are we adding the weight? It should be subtracted. Thanks.

"It's better to burn out then fade away" -DL 1983

 

[Chi Meson]

Well, it is added "as a vector," but in the algebraic equation it should, indeed, be subtracted. You can assign positives and negatives in any direction you like. the negatives are only directional indicators.

In equilibrium the net forces are zero, so T2Cos45+10N must be zero. Obviously, when written like this, T2Cos45 will turn out to be a negative quantity. I personally prefer to look at magnitudes in cases like this: T2Cos45 is balanced by 10N, so the magnitude is T2=10N/Cos45.

Similarly, Fy is zero, so T2Sin45 is balanced by N, so the magnitude of N=T2Sin 45 = [10N/Cos45]Sin45 =(10N)Tan45 = 10N

 

[blutoonwcarrotnnail]

T2Cos45+10N = 0 turning out to be negative is not the same thing as a directional indicator. The fact that 10N in the Fx component should be subtracted is something else. T2Cos45+10N is solving for the tension of T2 not setting the direction of 10N in Fx. The point is that if 10N is really supposed to be written T2Cos-10N = 0 then why is this against the laws of trigonometry? The later substitution of T2 Fx component into the T2 Fy component does not explain the trigonometric sign rule of why 10N is positive if it should be negative. Is Tan 45 the 45 degree angle which the 10N block makes with the equilibrium point or is it the 45 degree angle that
rope T2 makes to the x axis? Is this a shortcut to finding the solution? When presented with a right triangle the tangent of 45 degrees within the triangle itself of the original block will give you the answer? Thanks.

"It was the heat of the moment" -A 1982

 

[Chi Meson]

I read your last post three times. I have degrees in Physics and English. I do not know what you are saying.

 

[blutoonwcarrotnnail]

A ten Newton block is hanging from a line, goes up and over a pulley and to the right at a 90 degree angle. It hits an equilibrium point. At this eq point a line leaves upwards of tension T2 at 45degrees. Also at this eq point a string hangs down of unknown N.

Here are the components to find the second block:

Fx = T2Cos45 + 10N
Fy = T2Sin45 - N

Are these two components correct or should it be:

Fx = T2Cos45 - 10N
Fy = T2Sin45 + N

or should it be

Fx = T2Cos45 - 10N
Fy = T2Sin45 - N

Thanks.

"This is no time for foolin around" -Veronica L.