Question About Forces Using Unit Vectors

In summary: Question About Unit VectorsIn summary, the problem is that the student is unsure how to solve a problem that they have never seen before. They have identified the forces exerted on the top of the mast by the sailboat's forestay and backstay, and found that the total force is 180i- 820j. However, they are not sure how to find the magnitudes of the forces exerted at the location of the cables.
  • #1
imapeiceofwod
29
0
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
 
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  • #2
imapeiceofwod said:
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
i represents the unit vector in the horizontal direction and j represents the unit vector in the vertical direction. So the force is 180 N in the + i direction and 820 N in -j (or downward) direction.

First of all, draw a diagram of the mast and cables. Then draw a vector diagram of the forces of the cables (this is not a free-body diagram because we are not concerned with all the forces - the forces exerted by the mast and boat and gravity, for example). Write out the equations for downward and lateral force. To do this you must resolve the components of the cable tensions in the i and j directions.

AM
 
  • #3
Question About Unit Vectors

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?
 
  • #4


imapeiceofwod said:
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)
This is not correct.

Let the tension in AB be TAB and the components in the directions of the unit vectors be [itex]T_{ABi},T_{ABj}[/itex]. The following applies:

[tex]\vec{T_{AB}} = T_{ABi}\hat i + T_{ABj}\hat j[/tex]

[tex]T_{AB}^2 = T_{ABi}^2 + T_{ABj}^2[/tex]

AM
 
  • #5


I understand your confusion and am happy to help you with this problem. First, let's break down the given force into its components using unit vectors. 180i-820j (N) can be written as 180i + (-820)j (N). This means that the force in the x direction is 180 N and the force in the y direction is -820 N. Now, we can use the Pythagorean theorem to find the magnitude of the total force, which is the square root of (180^2 + (-820)^2) = 837.6 N.

Next, we need to find the individual forces exerted by the cables AB and BC at point B. To do this, we can use the fact that the total force is the sum of the individual forces. This means that the x component of the total force (180 N) must be equal to the sum of the x components of the individual forces exerted by AB and BC. Similarly, the y component of the total force (-820 N) must be equal to the sum of the y components of the individual forces exerted by AB and BC.

Now, let's look at the coordinates of points A, B, and C. We can see that AB and BC are both horizontal and vertical lines, which means that the forces exerted by these cables must also be horizontal and vertical. This means that the x components of the individual forces must be equal to the x coordinates of points A and C, and the y components of the individual forces must be equal to the y coordinates of points A and C.

Using this information, we can set up a system of equations:

180 = xA + xC
-820 = yA + yC

Substituting the coordinates of points A and C, we get:

180 = 0 + 9
-820 = 1.2 + 1

Solving for xA and yA, we get:

xA = 189 N
yA = -821.2 N

Similarly, for xC and yC, we get:

xC = -189 N
yC = 822.2 N

Therefore, the magnitudes of the individual forces exerted by the cables AB and BC at point B are 189 N and 822.2 N, respectively.

I hope this explanation helps you understand how to approach and solve this
 

1. How do unit vectors represent forces?

Unit vectors are used to represent forces by providing a direction and magnitude. They are typically denoted by the letters i, j, and k, and correspond to the x, y, and z axes respectively. By multiplying a unit vector by a scalar value, the resulting vector represents a force with the given magnitude and direction.

2. What is the significance of using unit vectors in force calculations?

Unit vectors are important in force calculations because they allow for a clear representation of direction and magnitude. By using unit vectors, we can easily break down a force into its components and accurately calculate its effect on an object's motion.

3. How do unit vectors relate to Newton's laws of motion?

Unit vectors are essential in applying Newton's laws of motion to real-world scenarios. They allow us to accurately calculate the forces acting on objects and determine their resulting acceleration and motion according to Newton's laws.

4. Can unit vectors be used for non-uniform forces?

Yes, unit vectors can be used for non-uniform forces. In these cases, the force is broken down into smaller components, and unit vectors are used to represent the direction and magnitude of each component. The resulting force vector is calculated by summing up all the individual force components.

5. How do unit vectors differ from regular vectors?

Unit vectors are different from regular vectors in that they have a magnitude of 1, whereas regular vectors can have any magnitude. Additionally, unit vectors are used to represent specific directions, whereas regular vectors can represent any direction in space.

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