Answer:
The 35N weight must be placed approximately 2.01 meters away from the pivot for the entire arrangement to be in equilibrium.
Step-by-step explanation:
For the system to be in equilibrium, the net torque about the pivot point must be zero. Torque is calculated as the product of the force and the perpendicular distance from the pivot.
Let's assume the pivot is at point O. The 20N weight is at a distance of 1.5m from O, the 45N weight is at a distance of 0.9m from O on the same side as the 20N weight, and the 35N weight is on the other side of O. Let's denote the distance of the 35N weight from O as "x."
The torque due to the 20N weight is 20N * 1.5m = 30 Nm (anticlockwise torque).
The torque due to the 45N weight is 45N * 0.9m = 40.5 Nm (anticlockwise torque).
The torque due to the 35N weight is 35N * x Nm (clockwise torque, since it's on the opposite side of the pivot).
For equilibrium, the sum of torques must be zero:
30 Nm + 40.5 Nm - 35N * x Nm = 0
Combine the torques:
70.5 Nm - 35N * x Nm = 0
Solve for x:
35N * x Nm = 70.5 Nm
x = 70.5 Nm / 35N
x = 2.01 meters
Therefore, the 35N weight must be placed approximately 2.01 meters away from the pivot for the entire arrangement to be in equilibrium.