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:
Explanation:
As we know that loop is placed in YZ plane and magnetic field is along x direction
So here net force on the side of the loop which lies along Y axis is given as
here we know that on Y axis z = 0
so B = 0
so we have
now on the opposite side we have z = a
so magnetic field is given as
so force on that side is given as
so net force on the loop is given as