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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?

  • physics -

    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).

  • physics -

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

    Fg = mg

  • physics -

    you just have to add w w w to the website

  • physics -

    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

  • physics -

    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.

  • physics -

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

  • Anonymous and bobpursley -

    can you please Anonymous and bobpursley explain this to me.

    Please and thank you

  • physics -

    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.

  • physics -

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

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