three horizontal forces are pulling on a ring at rest F1 is 100 N at a 45 degree angle and F2 is 135 N at a 160 degree direction what is the x component of F3
A sample of oxygen gas has a volume of 2.1 L at 3.6 atm pressure and 200 K temperature.
What would the volume of this sample be at 1.0 atm and 400 K?
how many liters ?
To find the x-component of F3, we need to break down F3 into its horizontal (x-axis) and vertical (y-axis) components.
We can start by finding the x-component of F1, which is F1x. The x-component can be found using the formula:
F1x = F1 * cos(theta1)
where F1 is the magnitude of the force and theta1 is the angle between the force and the x-axis.
Given that F1 is 100 N and the angle is 45 degrees, we can calculate the x-component of F1 as follows:
F1x = 100 N * cos(45 degrees)
Using the trigonometric identity cos(45 degrees) = √2 / 2, we can simplify the equation to:
F1x = 100 N * (√2 / 2)
F1x = 100 N * 0.707
F1x ≈ 70.7 N
Now, let's find the x-component of F2, which is F2x. We can use the same formula:
F2x = F2 * cos(theta2)
Given that F2 is 135 N and the angle is 160 degrees, we can calculate the x-component of F2 as follows:
F2x = 135 N * cos(160 degrees)
Using the trigonometric identity cos(160 degrees) = -0.9397 (rounded to four decimal places), we can simplify the equation to:
F2x = 135 N * (-0.9397)
F2x ≈ -126.95 N
Since the x-component of a force is a scalar quantity, we ignore the negative sign and consider the magnitude, so the x-component of F2 is approximately 126.95 N.
Finally, the x-component of F3 is simply the sum of the x-components of F1 and F2:
F3x = F1x + F2x
F3x ≈ 70.7 N + 126.95 N
F3x ≈ 197.65 N
Therefore, the x-component of F3 is approximately 197.65 N.
To find the x component of F3, we first need to break down the given forces into their x and y components.
Let's start with F1:
The magnitude of F1 is 100 N, and it is at a 45-degree angle. To find the x component of F1, we can use the formula:
Fx1 = F1 * cos(theta1),
where Fx1 is the x component of F1 and theta1 is the angle made by F1 with the x-axis.
Substituting the values, we get:
Fx1 = 100 N * cos(45 degrees)
Now, calculating Fx1:
Fx1 = 100 N * 0.707 ≈ 70.7 N (rounded to one decimal place)
Next, let's find the x component of F2:
The magnitude of F2 is 135 N, and it is at a 160-degree angle. Similarly, using the formula:
Fx2 = F2 * cos(theta2),
where Fx2 is the x component of F2 and theta2 is the angle made by F2 with the x-axis.
Substituting the values, we get:
Fx2 = 135 N * cos(160 degrees)
Now, calculating Fx2:
Fx2 = 135 N * (-0.9397) ≈ -126.9 N (rounded to one decimal place)
Finally, to find the x component of F3, we need to consider that the ring is at rest. This means that the sum of the x components of all forces acting on the ring must be zero.
Since F1 and F2 are pulling on the ring horizontally, the x component of F3 must be equal in magnitude but opposite in direction to the sum of Fx1 and Fx2.
Therefore, the x component of F3 is:
Fx3 = - Fx1 - Fx2
Substituting the values, we get:
Fx3 = - 70.7 N - (-126.9 N)
Now, calculating Fx3:
Fx3 = - 70.7 N + 126.9 N ≈ 56.2 N (rounded to one decimal place)
Hence, the approximate x-component of F3 is 56.2 N.
This is just like your other post. Just use x instead of y.
https://www.jiskha.com/questions/1796885/Three-horizontal-forces-are-pulling-on-a-ring-at-rest-f1-is-6-25-N-at-180-degree