Finding the Resultant by adding vectors. HELP

[oreocookie]

Finding the Resultant by adding vectors. HELP!

Homework Statement

1. Two forces act on an object. ONe has a magnitude of 166 Newtons and points at an angle of 60 degrees above the + X axis. The secon dhas a magnitude of 284 Newtons and points at an angle of 30 above the + X axis. A third force is applied and balances to zero effects of the other two. What are the magnitude and direction of this third force? Specify the direction relative to the negative x axis.

I'm ocnfused with this problem also...xD
2. What are the x and y components of the vector that must be added to the following three vectors so that the sum of hte four vectors is 0. Due east is the +x direction and due north is the +y direction.
A=133 units, 60 degrees south of west
B= 222 units, 35 degrees south of east
C= 177 units, 23 degrees north of east

Homework Equations

I am not really sure how to set up these two problems. I know what the variables and stuff are but what exactly am I looking for?...Plus, in problem number two, I don't understand the bolded statement.

The Attempt at a Solution

Problem 1) A=166 Angle A= 60 degrees B=284 Newtons Angle B=30.

Two forces act on an object. ONe has a magnitude of 166 Newtons and points at an angle of 60 degrees above the + X axis. The secon dhas a magnitude of 284 Newtons and points at an angle of 30 above the + X axis. A third force is applied and balances to zero effects of the other two. What are the magnitude and direction of this third force? Specify the direction relative to the negative x axis.

Problem 2) A=133 units, 240 degrees
B= 222 units, 325 degrees
C= 177 units, 23 degrees

 

[NBAJam100]

Homework Statement

1. Two forces act on an object. ONe has a magnitude of 166 Newtons and points at an angle of 60 degrees above the + X axis. The secon dhas a magnitude of 284 Newtons and points at an angle of 30 above the + X axis. A third force is applied and balances to zero effects of the other two. What are the magnitude and direction of this third force? Specify the direction relative to the negative x axis.

I'm ocnfused with this problem also...xD
2. What are the x and y components of the vector that must be added to the following three vectors so that the sum of hte four vectors is 0. Due east is the +x direction and due north is the +y direction.
A=133 units, 60 degrees south of west
B= 222 units, 35 degrees south of east
C= 177 units, 23 degrees north of east

Homework Equations

I am not really sure how to set up these two problems. I know what the variables and stuff are but what exactly am I looking for?...Plus, in problem number two, I don't understand the bolded statement.

The Attempt at a Solution

Problem 1) A=166 Angle A= 60 degrees B=284 Newtons Angle B=30.

Two forces act on an object. ONe has a magnitude of 166 Newtons and points at an angle of 60 degrees above the + X axis. The secon dhas a magnitude of 284 Newtons and points at an angle of 30 above the + X axis. A third force is applied and balances to zero effects of the other two. What are the magnitude and direction of this third force? Specify the direction relative to the negative x axis.

Problem 2) A=133 units, 240 degrees
B= 222 units, 325 degrees
C= 177 units, 23 degrees

Well there are a few ways you can solve this. It depends on how you want to. First, why don't you try to find the net force applied on the object... you can go about it a few ways. 1.) breaking the vectors into their X and Y components using trig and then adding them all up and see what the final net force would be and which direction it is in (keeping in mind that tan(angle)=Fy/Fx. or 2.) to once again find the total force on the object do... (Magnitude of vector)<cos(angle),sin(angle)>. then add them all up, and youll get a final force in the form of <x,y>. Find the magnitude by sqrt(x^2+y^2). to get total mag. Once again, tan(angle)=Fy/Fx.

 

[baba89]

To find the resultant force in this problem, we can use vector addition. This involves adding the x and y components of each vector separately. The x component is the magnitude of the vector multiplied by the cosine of the angle, and the y component is the magnitude multiplied by the sine of the angle.

For problem 1, we can draw the two given forces as vectors, with the first one at 60 degrees above the +x axis and the second one at 30 degrees above the +x axis. We can then use the Pythagorean theorem to find the magnitude of the third force, which will be the hypotenuse of a right triangle formed by the two given forces. The direction of the third force can be found using trigonometry, by taking the inverse tangent of the y component divided by the x component. This will give us the angle relative to the negative x axis.

For problem 2, we can use the given information to find the x and y components of each vector. For example, for vector A, the x component would be 133 units multiplied by the cosine of 60 degrees, and the y component would be 133 units multiplied by the sine of 60 degrees. We can then add all the x components together and all the y components together to find the resultant force in the x and y directions. Since the sum of all four vectors should be zero, we can set the resultant x and y components equal to 0, and solve for the missing component of the fourth vector. The direction of the fourth vector can be found using the same method as in problem 1.