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Question

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?

Answer 1

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).

Answer 2

y-direction: N1 + N2 - Fg = 0
x-direction: L1 x N1 - L2 x N2 = 0

Fg = mg

Answer 3

you just have to add w w w to the website

Answer 4

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

Answer 5

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.

Answer 6

okay but can you explain to me how to find N1 and N2

Answer 7

can you please Anonymous and bobpursley explain this to me.

Please and thank you

Answer 8

isolate N1 for the first eq:
N1 = Fg - N2

eq 2:
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.

Answer 9

Once the N1 is found, add the to Normal forces and voila!