Solving Unit Vectors Problem: Calculate Forces at B

[imapeiceofwod]

OK here's the problem. I am completely unsure how to solve it my teacher never even did an example like this before.

The total force exerted on the top of the mast B by the sailboat's forestay AB and backstay BC is 180i- 820j (N). What are the magnitudes of the forces exerted at B by the cables AB and BC?

A(0,1.2)
B(4,13)
C(9,1)

Any help please and thank you

here's what i got so far. I drew i diagram of the situation drawing and drew the vectors. I figured out Fab=(4i,11.8j) and Fbc=(5i,-12j)

Then i found the total force in the i direction which is 9i and the total force in the j which is 0.2j .

I know that the total force in the i direction is suppose to be 180i but i got 9i for the total force. Any help?

 

[pongo38]

You are confusing geometry with forces. There is a force triangle which is similar to a geometric triangle, and that property will enable you to solve the problem.
Where you have written Fab=... you should have written AB=(4i+11.8j). It's not a force, but the force in AB is a multiple of that vector. Can you work it from there?

 

[imapeiceofwod]

You are confusing geometry with forces. There is a force triangle which is similar to a geometric triangle, and that property will enable you to solve the problem.
Where you have written Fab=... you should have written AB=(4i+11.8j). It's not a force, but the force in AB is a multiple of that vector. Can you work it from there?

so would the force of AB be equal too square root 4^2 + 11.8^2? can you get me started a little bit more, other than that i could probely get it thank you

 

[pongo38]

No it's not sq root etc. You need to draw a two triangles in which the edges are parallel to AB, BC, and F= 180i- 820j. The first triangle expresses the geometry of the situation, and the second triangle is about forces. The two triangles are geometrically similar, and you should be able to solve the problem using that fact.

 

[rwisz]

I would approach this problem by first understanding the concept of unit vectors and how they can be used to represent forces in different directions. In this case, we are dealing with two forces, AB and BC, which are represented by the vectors Fab and Fbc, respectively. These vectors have both magnitude and direction, and they can be broken down into their respective components in the x and y directions.

To solve this problem, we need to use the concept of vector addition. The total force exerted at point B is the sum of these two forces, which can be represented by the vector Fb. This can be calculated by adding the x and y components of Fab and Fbc, giving us Fb = (9i, -0.2j).

Now, to find the magnitude of this force, we can use the Pythagorean theorem, which states that the magnitude of a vector is equal to the square root of the sum of the squares of its components. In this case, the magnitude of Fb is equal to the square root of (9^2 + (-0.2)^2), which is approximately 9.0 N.

To find the individual magnitudes of Fab and Fbc, we can use the same concept. The magnitude of Fab is equal to the square root of (4^2 + 11.8^2), which is approximately 12.4 N. Similarly, the magnitude of Fbc is equal to the square root of (5^2 + (-12)^2), which is approximately 13.0 N.

In summary, the magnitudes of the forces exerted at point B by the cables AB and BC are approximately 12.4 N and 13.0 N, respectively. It is important to note that the total force in the i direction is not equal to 180i, as this represents the x component of the total force. The total force in the i direction is actually 9i, as calculated earlier. I hope this explanation helps to clarify the solution to this problem.