1 Answer | Add Yours
Let the unit vector to the X direction is i and unit vector to the Y direction is j.
Let’s assume that the force 11N is along the X axis and force 7N is at the left side of 11N.
Let angles measure to counter clock wise directions are positive.
Now you will get the forces like this.
Force 11N is along X axis
Force 7N is 105 degree counter clock wise to for 11N
Force 15N is 147 degree clock wise to the force 11N
Now if you use vector notation you can write the forces as;
`11N = 11i+0j = 11 i`
`7N = 7*(cos105)i + 7*(cos(105-90))j = -1.8117 i + 6.7615 j`
`15N = 15*(cos(360-147)i+15*(cos(360-147-90) j =-12.5800 i -8.1696 j`
So resultant force `=11i - 1.8117 i + 6.7615 j - 12.5800 i - 8.1696 j`
`= -3.3917 i-1.4081 j `
Magnitude of the resultant force = `sqrt[(-3.3917)^2+(-1.4081)^2]`
= `3.6724 N`
Direction of the resultant force = `tan-1 ((-1.4081)/-3.3917)`
= `22.546 deg`
So the resultant force of 3.6724N is at 22.546 degrees counter clock wise angles to the 11N force.
We’ve answered 317,310 questions. We can answer yours, too.Ask a question