A uniform, 274-N rod that is 2.05m long carries a 225-N weight at its right end and an unknown weight W toward the left end.When W is placed 50.3cm from the left end of the rod, the system just balances horizontally when the fulcrum is located 78.0cm from the right end.
i tried using this formulae
w=lwo-2dwo-2d*225
i subbed in the values but my answer was wrong
To solve this problem, we can use the principle of moments or torque. Torque is the product of force and its perpendicular distance to the point of rotation.
Let's first find the torque produced by the 225-N weight at the right end. The distance of this weight from the fulcrum is 78.0 cm (0.78 m). Therefore, the torque produced by this weight is:
Torque_right = (force_right) * (distance_right) = (225 N) * (0.78 m)
Next, let's find the torque produced by the unknown weight, W, at the left end. The distance of this weight from the fulcrum is 50.3 cm (0.503 m). Therefore, the torque produced by this weight is:
Torque_left = (force_left) * (distance_left) = (W) * (0.503 m)
Since the system is balanced horizontally, the torques exerted by the weight on the right and the weight on the left must be equal:
Torque_right = Torque_left
Now we can set up the equation and solve for W, the weight at the left end:
(225 N) * (0.78 m) = W * (0.503 m)
Now let's solve this equation:
W = (225 N) * (0.78 m) / (0.503 m)
W ≈ 350 N