A box is pulled to the right with a force of 65 N at an angle of 58 degrees to the horizontal. The surface is frictionless. The
2 answers:
The free-body diagram is missing, but I assume the only forces acting on the box are the force F pushing the box, the weight of the object and the normal reaction of the surface.
Since the weight and the normal reaction acts in the vertical (y) direction, the only force acting on the box in the horizontal (x) direction is the horizontal component of the force F, which is given by
And so this is the net force in the x-direction.
Since the weight and the normal reaction acts in the vertical (y) direction, the only force acting on the box in the horizontal (x) direction is the horizontal component of the force F, which is given by
And so this is the net force in the x-direction.
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Answer: A symbolic expression for the net force on a third point charge +Q located along the y axis
Explanation:
Let the force on +Q charge y-axis due to +2Q charge be and force on +Q charge y axis due to -Q charge on x-axis be
.
Distance between the +2Q charge and +Q charge = d units
Distance between the -Q charge and +Q charge = units
= Coulomb constant
Net force on +Q charge on y-axis is:
The net froce on the +Q charge on y-axis is
