Need done as soon as possible please!!

tekilochka [14]

What you can do is multiply the whole first equation by 12, so 6x + 26.4y = 108.
then since the coefficients of the x value (the number in front of the x) are the same in both equations, you can subtract the second equation from the first:
6x + 26.4y = 108
-(6x + 0.4y  = -22)
   26y = 130 divide by 26
  y = 5
you can then plug y=5 into one of the equations to solve for x:
0.5x + 2.2(5) = 9
0.5x + 11 = 9 subtract 11 from both sides
0.5x = -2 multiply each side by 2
x = -4

5 0

4 years ago

Two forces with magnitudes of 100 and 50 pounds act on an object at angles of 50° and 160°, respectively. Find the direction and

MatroZZZ [7]

<h2>Hello!</h2>

Since there is no additional information, we can assume that the given angles are respect to the horizontal axis (x), so

The answer is:

Magnitude: 95.14 lbs

Direction: 80° respect to the x-axis

Since there is no additional information about the given angles, we can assume that both angles are respect to the horizontal axis (x), positive direction to x-axis positive directions (right) and above x-axis.

<h2>Calculations:</h2>

100 pounds (lb) force, 50° angle:

$100lbs*cos(50)=64.28lbs\\100lbs*sin(50)=76.60lbs$

50 pounds (lb) force, 160° angle:

$50lbs*cos(160)=-46.98lbs\\50lbs*sin(160)=17.10lbs$

So,

For x-axis forces we have:

64.28lbs (positive direction)

-46.98lbs (negative direction)

Then,

$64.28lbs-46.98lbs=17.3lbs$ (positive direction)

Fox y-axis forces we have:

76.60lbs (positive direction)

17.10 lbs (positive direction)

Then,

$76.60lbs + 17.10 lbs =93.7lbs$ (positive direction)

So, we have the two components values of the acting force:

17.3 lbs for the x-axis and 93.7 lbs for the y-axis.

Therefore,

We have to calculate the angle between both resultant force components:

Let's calll it α

So,

$\alpha=tan^{-1}(\frac{y-axisValue}{x-axisValue})=tan^{-1}(\frac{93.7}{17.3})\\\alpha=79.5=80$

Hence, the angle between both resultant force components is 80 °, meaning that the resultant force direction is positive, above the x-axis.

Now we have to calculate the magnitude of the resultant force since the sum of both resultant force components is equal to the hypotenuse of the formed triangle, we have that:

$sin(\alpha)=\frac{y-axisValue}{ResultantForceMagnitude}=\frac{93.7lbs}{ResultantForceMagnitude}\\\\ResultantForceMagnitude=\frac{93.7lbs}{sin(80)}=95.14lbs$