Determine the net result of a 35 N force and a 20 N force acting on an object at an angle of 125° to each other, using components.
I know that when I draw the vectors, I use parallelogram law to determine the complimentary angle, and I got 55degrees. However, on the answers, it says that the angle between the vector and its component was 35degrees. Where did this 35degrees come from?
To determine the net result of a 35 N force and a 20 N force acting on an object at an angle of 125° to each other using components, we can use vector addition techniques. Here's how you can find the correct angle:
1. Start by drawing a rough sketch of the vectors, making sure to indicate their magnitudes and directions. Label the 35 N force as F1 and the 20 N force as F2.
2. To find the components of each force, resolve them into their horizontal and vertical components. Use trigonometry to calculate these components.
Let's assume F1 makes an angle α with the x-axis, and F2 makes an angle β with the x-axis. The horizontal component of F1 is given by F1x = F1 * cos(α), and the vertical component of F1 is given by F1y = F1 * sin(α). Similarly, the horizontal component of F2 is F2x = F2 * cos(β), and the vertical component of F2 is F2y = F2 * sin(β).
3. Add the horizontal components of the forces together algebraically, and separately add the vertical components together. This will give you the total x-component (Fx) and the total y-component (Fy) of the net force.
Fx = F1x + F2x
Fy = F1y + F2y
4. Calculate the magnitude (F) of the net force using the Pythagorean theorem:
F = sqrt(Fx^2 + Fy^2)
5. Calculate the angle (θ) that the net force makes with the x-axis using trigonometry:
θ = arctan(Fy / Fx) (Note: be careful with your calculator settings to get the correct quadrant of the angle)
It's possible that the angle of 35 degrees mentioned in the answer refers to another angle in the problem, and not the one you're looking for. Double-check the context of where the 35-degree angle is mentioned to clarify its meaning in this situation.
<<the angle between the vector adn component>> which vector, which component?
Draw a polar chart, put the 35N force at 000, and the 20N @125.
Now call 000N, and 090 E
N vector components:
35+20cos125= ?23.5
E vector components
20 sin 125= ??16.4
Net result then is adding those two
magnitude:sqrt(23.5^2+16.4^2)=28.7
at an angle of arctan(16.4/23.5)=34.9 degrees from 090. I suspect that is where the 35 degrees came from, the true angle from 000 (N) would be 90-34.9 deg