# Applied Forces

1. Feb 6, 2006

### Huskies213

Can anyone help with what formulas to apply/use here ?

An object acted on by three forces moves with constant velocity. One force acting on the object is in the positive x direction and has a magnitude of 6.5 N; a second force has a magnitude of 3.5 N and points in the negative y direction. Find the direction ° (counterclockwise from theC +x axis) and magnitude (N) of the third force acting on the object.

2. Feb 6, 2006

### Staff: Mentor

The fact that the object moves with 'constant' velocity means that it is not accelerating, which would imply a change in velocity. The fact that it is not accelerating indicates that the net force or sum of forces is nil (0).

Therefore the third force must have a component equal to but opposite of the force in the +ve x-direction, and a y component which is equal to but oppositve the force in the -ve y-direction.

Then the angle is simply determined by an inverse trig function based on ratio of either the x or y component for the third force and the total or resultant force.

3. Feb 6, 2006

### Huskies213

Re

i got the answer of 7.38 for the first part (which is correct) but i'm still unsure about the second part. can anyone help ?

4. Feb 6, 2006

### Staff: Mentor

The vector is in the upper left quadrant (of the Cartesian coordinate system) since its x-component must be - to offset the + x-force, and its y-component must be + to offset the - y-force.

7.38 is correct! This should simply be $\sqrt{3.5^2\,+\,6.2^2}$.

Think of a vector F which has two components, Fx and Fy, and both components are orthogonal.

The magnitude of F = $\sqrt{(F_x^2\,+\,F_y^2)}$, and Fx = F cos$\theta$ and Fy = F sin$\theta$, where $\theta$ is the angle between F and the x-axis.

Think of the definition of the cos of an angle and how it relates to the legs and hypotenuse of a right triangle.

Last edited: Feb 6, 2006
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