Finding direction of force vector

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Erenjaeger
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Homework Statement


The magnitude of a force vector is 80.6 Newtons (N). The x component of this vector is directed along the +x axis and has a magnitude of 78.0 N. The y component points along the +y axis. (a) Find the angle between and the +x axis. (b) Find the component of along the +y axis.[/B]

Homework Equations

[/B]

The Attempt at a Solution


I tried using the x and y components as distances on an x y graph and then finding the resultant force vector thinking it could be the hypotenuse, and then finding the angle between the bottom of the hypotenuse which was 78 units along the x-axis and then at that corner finding the angle which i used CAH of sohcahtoa to find cos(theta) as a/h or 78.043/2.6 giving me 30.016. [/B]
 
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Erenjaeger said:
1. Homework Statement
The magnitude of a force vector https://edugen.wileyplus.com/edugen/courses/crs7924/art/qb/qu/c01/EAT_1319521635256_0_8466828615424514.gif is 80.6 Newtons (N). The x component of this vector is directed along the +x axis and has a magnitude of 78.0 N. The y component points along the +y axis. (a) Find the angle between https://edugen.wileyplus.com/edugen/courses/crs7924/art/qb/qu/c01/EAT_1319521635256_0_16365013397000572.gif and the +x axis. (b) Find the component of https://edugen.wileyplus.com/edugen/courses/crs7924/art/qb/qu/c01/EAT_1319521635257_0_1791308789505699.gif along the +y axis.2. Homework Equations 3. The Attempt at a Solution
It sounds as though there asking you to use trig to solve thing question, but I am unsure on how to go about it.
Don't know what the crosses stand for.
The relevant equations have disappeared and so has your attempt at solution. Something about the sound of the exercise has appeared, but exercises don't make sound.
Please correct :rolleyes: and hope the mentors don't delete this outright ...
 
BvU said:
Don't know what the crosses stand for.
The relevant equations have disappeared and so has your attempt at solution. Something about the sound of the exercise has appeared, but exercises don't make sound.
Please correct :rolleyes: and hope the mentors don't delete this outright ...
done
 
Good !
Erenjaeger said:
I tried using the x and y components as distances on an x y graph and then finding the resultant force vector thinking it could be the hypotenuse, and then finding the angle between the bottom of the hypotenuse which was 78 units along the x-axis and then at that corner finding the angle which i used CAH of sohcahtoa to find cos(theta) as a/h or 78.043/2.6 giving me 30.016, not sure if this is correct or not
You only have the x-component (the 78 N). Call the y component ##F_y##. What would be the total force ? The total force is given as 80.6, so you can write 80.6 = ...(something with ##F_y## and the 78 N) ...
Make a sketch (you probably did, but post it -- or describe it. I find "CAH of sohcahtoa" very original, but can't make sense of it.