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!
- 1 decade agoFavorite Answer
| T1 T2 |
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
- 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)