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Mei asked in Science & MathematicsPhysics · 3 years ago

PHYSICS HELP! (Will award best answer!)?

A 150 kg mass hangs from the end of a 7.00 m -long boom (rod) which has a mass of 65.0 kg. The boom is kept in position by a restraining cable attached 3/4 of the way

along it.

(a) Draw a free-body diagram of the boom

indicating all of the forces acting on it in

their correct locations.

(b) What is the magnitude of the tension in

the cable?

(c) What is the magnitude and direction of

the force of the hinge on the boom?

Please explain how you got the answers!

Attachment image

3 Answers

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  • Anonymous
    3 years ago
    Favourite answer

    Part a)

    The forces acting on the boom are its own weight (Mg) originating from the center of mass

    at half its length. The weight (mg) also acts on it together with the tension and also the

    hinge has a counterforce. These can all be separated into components and unfortunately

    I have no possibility to provide a picture.

    Part b)

    Torque: τ = rF

    We can choose the hinge as the reference and balance torques about that point.

    Sum of all torques is zero since the boom is held in equilibrium:

    Let clockwise be the positive direction of torque, ∑τ+ = 0 =>

    (½L)Mgcos50° + (L)mgcos50° – (¾L)Tsin70° = 0

    The boom's length L can be crossed on both sides and won't be needed anymore.

    So this rearranges to:

    gcos50°(½M+m) = ¾Tsin70°

    T = 4gcos50°(½M+m) / 3sin70°

    Acceleration of gravity g=9.8m/s² and you also have M=65kg and m=150kg. Plug in and calculate:

    T = 1631 N

    (≈ 1.6 kN to two significant figures)

    Part c)

    The force of the hinge countering the forces from the cable, mass and boom whose components

    are directed along the boom.

    Fhinge = gsin50°(M+m) + Tcos70°

    Fhinge = 2172 N

    or 2.2 kN to two significant figures, directed at 50degrees, upwards along the boom

  • 3 years ago

    To determine the tension in the cable, let’s do a torque problem. Let the pivot point be at the hinge. The tension in the cable produces counter clockwise torque. The weights of the box and the boom produce clockwise torque.

    ¾ * 7 = 5.25 meter

    Counter clockwise torque = T * 5.25 * sin 70

    Clockwise torque = 150 * 9.8 * 7 *cos 50 = 10,290 * cos 50

    Clockwise torque = 65 * 9.8 * 3.5 * cos 50 = 2,229.5 * cos 50

    Total clockwise torque = 12,519.5 * cos 50

    T * 5.25 * sin 70 = 12,519.5 * cos 50

    T = 12,519.5 * cos 50 ÷ (5.25 * sin 70) = 2384⅔ * cos 50 ÷ sin 70

    This is approximately 1631 N.

  • oubaas
    Lv 7
    3 years ago

    cos 50° = 0.643

    ((65*7/2+150*7)*cos 50°)*g = (Ty*7*3/4*cos 50)

    Ty = (65*7/2+150*7)*g*4/21 = 243.33 g N

    T = Ty/(sen (180-(90+70)) = 243.33/0.342*g = 711.5*g N (≈ 6977 N)

    Fy = m*g = (150+65)*g N

    Fx = T*cos 20°= 6977*0.94 = 6558 N

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