Vector word problems?

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March 27th 2011, 08:39 AM
homeylova223

Vector word problems?

What is the magnitude and direction of the resultant of a 105 newton force along the x axis and a 110 newton force at an angle 50 degree to one another?

Find the magnitude and direction of the resultant of two force of 250 pounds and 45 pounds at angles 25 degrees and 250 degrees with the x axis respectively?

Can anyone show me how to these word problems?
March 27th 2011, 10:17 AM
skeeter

Quote:

Originally Posted by homeylova223

What is the magnitude and direction of the resultant of a 105 newton force along the x axis and a 110 newton force at an angle 50 degree to one another?

Find the magnitude and direction of the resultant of two force of 250 pounds and 45 pounds at angles 25 degrees and 250 degrees with the x axis respectively?

Can anyone show me how to these word problems?

first of all, draw a sketch of the vectors to be added.

there are different methods, but I find using the method of components to be fairly simple ...

$\displaystyle \sum{F_x} = 105 + 110\cos(50^\circ)$

$\displaystyle \sum{F_y} = 0 + 110\sin(50^\circ)$

$|F_R| = \sqrt{(F_x)^2 + (F_y)^2}<br />$

direction of the resultant , $\theta = \arctan\left(\dfrac{F_y}{F_x}\right)$

... try same method with the second problem
March 27th 2011, 02:11 PM
HallsofIvy

Quote:

Originally Posted by homeylova223

What is the magnitude and direction of the resultant of a 105 newton force along the x axis and a 110 newton force at an angle 50 degree to one another?

I interpret "at an angle 50 degree to one another" to mean that the 110 N forces is at 50 degrees to the 105 N force. That is NOT the way skeeter interpeted it- he interpreted this to mean that the 110 N force is at 50 degrees to the x-axis so that the angle between the two vectors is 180- 50= 130 degrees.

As skeeter said, there are many ways to do these problems. One way is to use the "cosine law" for general triangles. If my interpretation is correct, then the resultant force is the side of a triangle opposite the 50 degree angle between two sides of length 110 and 105. By the cosine law, the length of that third side is given by $c^2= (110)^2+ (105)^2- 2(110)(105)cos(50)$ . Once you know that, You can use the cosine law again to find the angle opposite the 105 N force, the angle the resultant makes with the x- axis.

Quote:

Find the magnitude and direction of the resultant of two force of 250 pounds and 45 pounds at angles 25 degrees and 250 degrees with the x axis respectively?

Here you specifically say that the two forces make angles of 25 degrees and 250 degrees with the x-axis. That means that the angle between the two vectors is 45 degrees. Now you can use the cosine law to find the length of the resultant as $c^2= (250)^2+ (250)^2- 2(250)(250)cos(45)$ .

Quote:

Can anyone show me how to these word problems?
March 27th 2011, 04:45 PM
homeylova223

Skeeter interpreted correctly.
March 28th 2011, 04:35 PM
homeylova223

Find the magnitude and direction of the resultant of two force of 250 pounds and 45 pounds at angles 25 degrees and 250 degrees with the x axis respectively?

I am still having problems doing this one I mean I think I know what I would use for x axis FX= 250+45 cos (250) But what would I do for y axis?
March 28th 2011, 05:06 PM
skeeter

Quote:

Originally Posted by homeylova223

Find the magnitude and direction of the resultant of two force of 250 pounds and 45 pounds at angles 25 degrees and 250 degrees with the x axis respectively?

I am still having problems doing this one I mean I think I know what I would use for x axis FX= 250+45 cos (250) But what would I do for y axis?

just use the x-axis ...

$\displaystyle \sum{F_x} = 250\cos(25) + 45\cos(250)<br />$

$\displaystyle \sum{F_y} = 250\sin(25) + 45\sin(250)<br />$