posted by anna .
this is basically the diagram to my question
A person of mass m = 62 kg is doing push-ups as shown in the attached figure. The distances are a = 91 cm and b = 60% of l1. Calculate the vertical component of the normal force exerted by the floor on both hands.
Can someone please explain this problem to me i don't know how to approach it..i understand that i will have to use times each distance by the weight but i don't understand what to do next?
I cant load the site, I will try later. The idea is that you sum moments (clockwise+, counterclockwise-) and set equal to zero. A moment is Force*distance*sinAnglebetweenthem(usually 90 degrees).
y-direction: N1 + N2 - Fg = 0
x-direction: L1 x N1 - L2 x N2 = 0
Fg = mg
you just have to add w w w to the website
can u explain to me what you mean by y direction is equal to and for the force *distance *the angle i don't know the angle for the question
Since we're only trying to find the Normal Force exerted by the two hands, we automatically assume that it's in mechanical equilibrium, thus equaling to zero.
okay but can you explain to me how to find N1 and N2
can you please Anonymous and bobpursley explain this to me.
Please and thank you
isolate N1 for the first eq:
N1 = Fg - N2
N1 x L1 - N2 x L2 = 0
sub N1 from eq 1 to eq 2.
(Fg - N2)L2 - N2 x L2=0
(Fg - N2 x L2) - N2 x L2 = 0
Fg = N2 x L2 + N2 x L1
Fg = N2(L2+L1)
since we already know Fg = mg, we isolate for N2
N2 = Fg/(L2+L1)
Thus the normal for N2 is found. Now look back to eq 1 again:
N1 + N2 - Fg = 0
then isolate for N1 by substituting the found value N2.
Make sure you convert cm to m.
Once the N1 is found, add the to Normal forces and voila!