Two students, Jahna, and Raul, re performing Experiment 1. To simplify their calculations, each will apply their force perpendicular to the arm of the apparatus. Jahna attaches her spring scale 15cm from the pivot and pulls with a force of 2.5N. Raul attaches his springs scale 10cm from the pivot and pulls so that the apparatus is in equilibrium. What is the reading on Raul's spring scale?
15 * 2.5 = 10 * X
solve for X
Thank You!
To solve this problem, we can use the concept of torque. Torque is the rotational equivalent of force and is calculated by multiplying the force by the perpendicular distance from the point of rotation.
Let's calculate the torque exerted by Jahna on the apparatus. Jahna applies a force of 2.5N at a perpendicular distance of 15cm from the pivot. Converting the distance to meters, we have 15cm = 0.15m. Therefore, the torque exerted by Jahna is:
Torque by Jahna = Force x Perpendicular Distance
= 2.5N x 0.15m
= 0.375 Nm
For the apparatus to be in equilibrium, the torques exerted by Jahna and Raul must cancel each other out. Since Raul's spring scale is attached 10cm (0.1m) from the pivot, we can calculate the reading on Raul's spring scale using the torque equation:
Torque by Raul = Torque by Jahna
Force by Raul x Perpendicular Distance by Raul = 0.375 Nm
Let's denote the force on Raul's spring scale as F_Raul. Hence, we have:
F_Raul x 0.1m = 0.375 Nm
Solving for F_Raul:
F_Raul = 0.375 Nm / 0.1m
= 3.75 N
Therefore, the reading on Raul's spring scale is 3.75 N.