To practice Problem-Solving Strategy 11.1 Equilibrium of a Rigid Body. A horizontal uniform bar of mass 2.7 kg and length 3.0 m is hung hori

Question

To practice Problem-Solving Strategy 11.1 Equilibrium of a Rigid Body. A horizontal uniform bar of mass 2.7 kg and length 3.0 m is hung horizontally on two vertical strings. String 1 is attached to the end of the bar, and string 2 is attached a distance 0.6 m from the other end. A monkey of mass 1.35 kg walks from one end of the bar to the other. Find the tension T1 in string 1 at the moment that the monkey is halfway between the ends of the bar.

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Euphemia 4 years 2021-08-26T03:58:47+00:00 1 Answers 0 views 0

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  1. Nick
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    2021-08-26T04:00:03+00:00
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    Answer:

    T_{1} = 14.88 N

    Explanation:

    Let’s begin by listing out the given variables:

    M = 2.7 kg, L = 3 m, m = 1.35 kg, d = 0.6 m,

    g = 9.8 m/s²

    At equilibrium, the sum of all external torque acting on an object equals zero

    τ(net) = 0

    Taking moment about T_{1} we have:

    (M + m) g * 0.5L – T_{2}(L – d) = 0

    ⇒ T_{2} = [(M + m) g * 0.5L] ÷ (L – d)

    T_{2} = [(2.7 + 1.35) * 9.8 * 0.5(3)] ÷ (3 – 0.6)

    T_{2}= 59.535 ÷ 2.4

    T_{2} = 24.80625 N ≈ 24.81 N

    Weight of bar(W) = M * g = 2.7 * 9.8 = 26.46 N

    Weight of monkey(w) = m * g = 1.35 * 9.8 = 13.23 N

    Using sum of equilibrium in the vertical direction, we have:

    T_{1} + T_{2} = W + w   ——- Eqn 1

    Substituting T2, W & w into the Eqn 1

    T_{1} + 24.81 = 26.46 + 13.23

    T_{1} = 14.88 N

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