Boat pulled by 2 Ropes, Find Magnitude of Force in Each

[crybloodwing]

So I am given the following diagram.

I labeled the top rope X and the bottom rope Y.

I need to find the magnitude of the force in rope X and in rope Y.

So I know that X+Y=455N.



I tried using SOHCAHTOA to figure out the lengths/magnitudes. I put the vectors together to make one triangle.



X would be Cos(20)=Adjacent/Hypo, or X/455. So
X= 455*Cos(20)
X=427.5601

So then if I did 455-X, I would get Y=27.4398

However, if I get Y using SOHCAHTOA:
sin(20)=y/455
y=455*sin(20)
y=155.619

Then Y+X does not equal 455. It then equals 583.1791.

What am I doing wrong here?

 

[sinx]

So I am given the following diagram.

I labeled the top rope X and the bottom rope Y.

I need to find the magnitude of the force in rope X and in rope Y.

So I know that X+Y=455N.

View attachment 10249

I tried using SOHCAHTOA to figure out the lengths/magnitudes. I put the vectors together to make one triangle.

View attachment 10250

X would be Cos(20)=Adjacent/Hypo, or X/455. So
X= 455*Cos(20)
X=427.5601

So then if I did 455-X, I would get Y=27.4398

However, if I get Y using SOHCAHTOA:
sin(20)=y/455
y=455*sin(20)
y=155.619

Then Y+X does not equal 455. It then equals 583.1791.

What am I doing wrong here?

xcos20+ycos30=455

 

[Dr.Peterson]

So I am given the following diagram.

I labeled the top rope X and the bottom rope Y.

I need to find the magnitude of the force in rope X and in rope Y.

So I know that X+Y=455N.

View attachment 10249

I tried using SOHCAHTOA to figure out the lengths/magnitudes. I put the vectors together to make one triangle.

View attachment 10250

X would be Cos(20)=Adjacent/Hypo, or X/455. So
X= 455*Cos(20)
X=427.5601

So then if I did 455-X, I would get Y=27.4398

However, if I get Y using SOHCAHTOA:
sin(20)=y/455
y=455*sin(20)
y=155.619

Then Y+X does not equal 455. It then equals 583.1791.

What am I doing wrong here?

When you say, "X would be Cos(20)=Adjacent/Hypo, or X/455", you appear to be assuming the angle at the top of your triangle is a right angle. It isn't. So your calculation of X is wrong. The same is true of your calculation of Y.

Didn't you wonder why you didn't use the 30 degree angle at all?

 

[Dr.Peterson]

xcos20+ycos30=455

That's one equation you need, using components in the direction of the resultant. The other equation will be based on the perpendicular components.

 

[Harry_the_cat]

So I am given the following diagram.

I labeled the top rope X and the bottom rope Y.

I need to find the magnitude of the force in rope X and in rope Y.

So I know that X+Y=455N.

View attachment 10249

I tried using SOHCAHTOA to figure out the lengths/magnitudes. I put the vectors together to make one triangle.

View attachment 10250

X would be Cos(20)=Adjacent/Hypo, or X/455. So
X= 455*Cos(20)
X=427.5601

So then if I did 455-X, I would get Y=27.4398

However, if I get Y using SOHCAHTOA:
sin(20)=y/455
y=455*sin(20)
y=155.619

Then Y+X does not equal 455. It then equals 583.1791.

What am I doing wrong here?

It is not a right angled triangle so SOHCAHTOA does not apply. You could look at the horizontal and vertical components of the vectors as others have suggested.

You can solve it using your triangle and the law of sines (if you have learnt that). You need to look carefully at where the 30 degree angle comes into your triangle diagram. Can you see that the angle in the bottom right corner of your triangle is 30 degrees? The you can calculate the other angle and the lengths of the two sides.

 

[crybloodwing]

It is not a right angled triangle so SOHCAHTOA does not apply. You could look at the horizontal and vertical components of the vectors as others have suggested.

You can solve it using your triangle and the law of sines (if you have learnt that). You need to look carefully at where the 30 degree angle comes into your triangle diagram. Can you see that the angle in the bottom right corner of your triangle is 30 degrees? The you can calculate the other angle and the lengths of the two sides.

So I have angles of 20, 30, and 130. I used the law of sins a/sin A = b/sin B = c/sin C. so 455/sin 130 = x/sin 30 = y/sin 20

I got 455/sin 130= 593.96

So x= 296.98 and y= 203.14639

But that equals 500.1265, not 455....

 

[crybloodwing]

xcos20+ycos30=455

I do understand that, but I am unsure how to find X and Y...

 

[sinx]

I do understand that, but I am unsure how to find X and Y...

you have 2 unknowns, x and y.
so you need 2 eqns.
1) xcos20+ycos30=455
2) xsin20+ysin30=0 (it is not moving (and/or forces balance) in the vertical direction)