- #1
ballooza
- 14
- 0
if the normal force defined as the force that acts on an object when it touches the surface perpendicularly, wouldn't it be at the opposite direction of the Weight (W) as a vector and with the same absolute value.. so that the formula would be : F(n)= -mg
if this formula is correct (which is surely), so how come it's being calculated like this:
the normal force acting on a 70 kg-person would be
FN = - (70 kg)(-9.8 m/s2) = 686 N
while the weight of the same person is:
F=mg
= 70 * 9.8 = 686 N
1)shouldn't be the weight positive while the normal force is negative?
2) could i be scientifically correct, if i say that the normal force is considered the Re-action force due to the act of the weight of the body (u know... body acts force on surface while surface acts force on body)
while action and reaction forces should be applied to one single point??
background: when we start solving mechanics problems we proof that a body is stable when the net force= zero by formula: net force= F(n)+W+F+F(friction)
so how come we can add F(n)+W together while they don't apply thy forces to the same one point?
if this formula is correct (which is surely), so how come it's being calculated like this:
the normal force acting on a 70 kg-person would be
FN = - (70 kg)(-9.8 m/s2) = 686 N
while the weight of the same person is:
F=mg
= 70 * 9.8 = 686 N
1)shouldn't be the weight positive while the normal force is negative?
2) could i be scientifically correct, if i say that the normal force is considered the Re-action force due to the act of the weight of the body (u know... body acts force on surface while surface acts force on body)
while action and reaction forces should be applied to one single point??
background: when we start solving mechanics problems we proof that a body is stable when the net force= zero by formula: net force= F(n)+W+F+F(friction)
so how come we can add F(n)+W together while they don't apply thy forces to the same one point?