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Anonymous asked in Science & MathematicsPhysics · 1 decade ago

Tension force problem...please help?

A 20 m long beam with a mass 40 kg is supported by a rope at each end. A 30 kg weight is 4 m from one end. Find the tension in each rope. Please help me with this! I don't understand it and the examples in my book aren't helping!

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  • 1 decade ago
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    | |

    | T1 T2 |

    | |

    A ---------------------|--------|-------------B

    | |

    40*9.8 =392 N | | 30 *9.8 = 294 N

    apply torque at point A(left end)

    20(T2) = 392(10) + 294(16) solving for T2 you get

    T2= 431.2 N

    using sum of forces in the y axis you get

    T1 + T2 = 392 + 294

    T1 = 254.8 N

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  • 1 decade ago

    Weights are acting downward.

    The total weight acting down = 40 g +30g = 70g.

    Since the beam is at rest, the upward force = downward force.

    If P and Q are the tensions in the rope, the upward force = P + Q.

    P + Q = 70 g.

    Though they are balanced the may rotate about a point if there is net torque on the system.

    For equilibrium the net torque about any point must also be zero.

    Torque is force x perpendicular distance.

    Let us find the torque or moment about one end say where one rope is tied.

    The weight of the beam acts at its center, (10 m from one end.

    Its moment about the end, where P (say in the left) acts, is 40g x 10.

    This tends the beam to rotate in the clockwise direction.

    The moment of the mass 30 kg is 30g x 4 (since it acts at 4 m from the left)

    This force also rotates the beam in clockwise direction,

    The moment of the force Q acting at the extreme right is Q x 20. This rotates the beam in the anticlockwise direction since it acts upwards.

    The moment of the force P is zero since the distance is zero.

    For equilibrium the anticlockwise moment = clockwise moment.

    Q x 20 = 40g x 10 + 30g x 4 = 520 g

    Q = 26 g N or 26 kg force. (254.8 N)

    Since P + Q = 70 g,

    P = 70 -26 =44 g or 44 kg force. (431.2 N)

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