# 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!

### 2 Answers

- 1 decade agoFavorite Answer
| |

| T1 T2 |

| |

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

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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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- PearlsawmeLv 71 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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