A particle is held in equilibrium by three coplanar forces as shown in the diagram below:
The three forces are:
- A force of magnitude 13 N acting at an angle θ \theta\,θ above the horizontal, where tanθ=125\displaystyle \tan \theta = \frac{12}{5}tanθ=512
- A horizontal force of magnitude F F\,F acting to the left
- A vertical force of magnitude T T\,T acting downwards
If the force of magnitude F F\,F is removed, what is the magnitude and direction of the resultant force now acting on the particle?
12 N12\text{ N}12 N horizontally to the right
5 N5\text{ N}5 N horizontally to the left
5 N5\text{ N}5 N horizontally to the right
12 N12\text{ N}12 N vertically downwards