physics forces problem...?

When a man stands on a scale, it reads 92.6kg. When he pulls down on the chin-up bar installed over the scale, the reading decreases to 75.1kg.

What is the magnitude of the force he exerts on the chin-up bar?

Comments

  • 175 Newtons

    its all about normal force

    http://www.lockershelfco.com/

  • Sum Fx = zero and Sum Fy = zero Sum Fx = F1,x + F2,x + F3,x = zero + forty four cos 60 + F3,x = 22 + F3,x = zero. So, F3,x = -22 N. Sum Fy = F1,y + F2,y + F3,y = 33 + forty four sin 60 + F3,y = 33 + forty four sin 60 + F3,y = zero So, F3,y = -seventy one N. F3 = sqrt [ F3,x ^two + F3,y ^two ] = sqrt [ (-22)^two + (-seventy one)^two ] = seventy four N. Direction of F3 = a hundred and eighty + tan^(-one million) (seventy one/22) = a hundred and eighty + seventy three = 253 deg.

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