Regarding vectors, resultants, and angles

[ble]

I have two problems that I cannot seem to figure out. I already know the answers, but I want to know the process of how to complete them so I know for quizzes and tests.

1. Vector A is 3 units in length and points along the positive x-axis; vector B is 4 units in length and points along a direction 150 degrees from the positive x-axis. What is the direction of the resultant with respect to the positive x-axis?

The correct answer is 103 degrees, but I want to know how to get to it.

2. The following force vectors act on an object: i.) 50 Newtons at 45 degrees north of east and ii.) 25 Newtons at 30 degrees south of east. Which of the following represents the magnitude of the resultant and its angle relative to the easterly direction?

The correct answer is 61.4 Newtons at 21.8 degrees, but again, I want to know how to get to it.

 

[Greg Bernhardt]

Show us some effort and the relevant equations.

 

[phinds]

and draw the vector diagram

 

[BvU]

Hello ble, and welcome to PF. :-)

The number 2. in the template is there so that potential helpers can provide assistance at an adequate level. After all, that's literally what you are asking for. The 3 is to show what you have done so far, so you can get help at the point where you are stuck. Much better than starting with nothing.

So don't discard the template but make good use of it. Its use happens to be compulsory in PF.

 

[HalfLostAndSearching]

For the first problem, you can use the trigonometric definition of a vector to find the magnitude and direction of the resultant. The magnitude of the resultant is given by the formula R = √(A^2 + B^2 + 2ABcosθ), where A and B are the magnitudes of the two vectors and θ is the angle between them. In this case, A = 3 units and B = 4 units, and the angle between them is 150 degrees. Plugging these values into the formula, we get R = √(3^2 + 4^2 + 2(3)(4)cos150) = √(9 + 16 + 24(-0.866)) = √(25 - 20.784) = √4.216 = 2.053 units.

The direction of the resultant can be found using the inverse tangent function. The direction, θr, is given by the formula θr = tan^-1(Bsinθ/(A + Bcosθ)). In this case, Bsinθ = 4sin150 = 2 units and A + Bcosθ = 3 + 4cos150 = 3 + 4(-0.866) = 0.604 units. Plugging these values into the formula, we get θr = tan^-1(2/0.604) = tan^-1(3.311) = 103 degrees.

For the second problem, you can use the parallelogram law of vector addition to find the resultant. The magnitude of the resultant is given by the formula R = √(A^2 + B^2 + 2ABcosθ), where A and B are the magnitudes of the two vectors and θ is the angle between them. In this case, A = 50 Newtons and B = 25 Newtons, and the angle between them is 30 + 45 = 75 degrees. Plugging these values into the formula, we get R = √(50^2 + 25^2 + 2(50)(25)cos75) = √(2500 + 625 + 2500(0.259)) = √(3125 + 650) = √3775 = 61.4 Newtons.

The direction of the resultant can be found using the inverse tangent function