Solving Three Force Motion Problem: Find Third Force

[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.

 

[Astronuc]

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.

 

[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 ?

 

[Astronuc]

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.

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, F_x and F_y, and both components are orthogonal.

The magnitude of F = $\sqrt{(F_x^2\,+\,F_y^2)}$, and F_x = F cos$\theta$ and F_y = 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.