Vertically, the object is in equilibrium, so that the net force in this direction is
∑ <em>F</em> (vertical) = <em>n</em> - <em>mg</em> = 0
where <em>n</em> is the magnitude of the normal force due to the contact between the object and surface. You're given that the object's weight is <em>mg</em> = 550 N, so <em>n</em> = 550 N as well.
Horizontally, the net force would be
∑ <em>F</em> (horizontal) = <em>p</em> - <em>f</em> = 0
where <em>p</em> is the magnitude of the applied force and <em>f</em> is the magnitude of (kinetic) friction opposing <em>p</em>. Now,
<em>f</em> = 0.012<em>n</em> = 0.012 (550 N) = 6.6 N
so that you need to apply a force of <em>p</em> = 6.6 N to keep the object sliding at a steady pace.
Answer:
Explanation:
As we know that the force equation at the top is given as
now we know that
so we have
so we will have
Answer: B. A hearing by a federal court of appeals
