100 kg jeepney is at reast on a bridge 5m from the right end, neglecting the weight of the bridge. find f1 and f2

To find the forces F1 and F2 acting on the jeepney, we need to consider the forces acting on it.

Let's assume F1 is the force acting on the jeepney from the left side of the bridge, and F2 is the force acting on the jeepney from the right side of the bridge.

Since the jeepney is at rest and neglecting the weight of the bridge, the forces must be balanced. This means that the sum of the forces acting on the jeepney in the horizontal direction must be equal to zero.

To determine the forces, we need to consider the lever arm and torque.

Torque is a measure of the turning force on an object. It is calculated by multiplying the force applied to an object by the lever arm distance:

Torque = Force x Lever arm

In this case, we can consider the pivot point as the left end of the bridge, and the torque due to F1 is F1 x 5m (lever arm).
Similarly, the torque due to F2 is F2 x 0m (as it is acting at the right end of the bridge).

Since the jeepney is at rest, the torques must also be balanced. Therefore, the torque due to F1 must be equal in magnitude but opposite in direction to the torque due to F2.

So, we can write the equation:

F1 x 5m = -F2 x 0m

Since F2 x 0m = 0, the equation becomes:

F1 x 5m = 0

Dividing both sides by 5m:

F1 = 0

This means that F1 is zero, indicating there is no force acting on the jeepney from the left side of the bridge.

Now, to find F2, we need to consider the weight of the jeepney. Let's assume the weight of the jeepney is 980 N (approximately 100 kg * 9.8 m/s^2, taking the acceleration due to gravity as 9.8 m/s^2).

Since the jeepney is at rest, the force due to the weight must be balanced by an equal and opposite force (F2) acting upward:

F2 = 980 N

Therefore, F1 = 0 N and F2 = 980 N.