Question

Can you help finish drawing the free body diagram for question #5?

Also I need help with #6 for calculating torque and filling in the table.

I will rate thank you!

Name: Balance Lab (Part 1) Date: Explore the "Balance Lab" section of the Balancing Act simulation. 4. Create a situation in which the beam is balanced (without the supports) while there is a single collection of bricks on one side of the pivot and two different collections of bricks on the other side (3 collections of bricks in total in the experiment). Draw the situation on the figure below while making sure to label the masses and the distance to each mass from the center (pivot point). 15kg lokg 5kg 5kg 5kg 2 1.75 1.5 1.25 / 75.5.257 125.575 1 1.25 15 1,5 2 V | v 5. Turn the picture above into an extended free-body diagram. In the picture above, draw in the forces pushing downward on the beam due to the weight of each collection of bricks from each mass. Label the forces with the symbols wi, w, and ws. (Note that there is a checkbox in the program for “Forces from Objects” that can help you with this step.) Next, draw the weight of the beam at its "center of mass” which should be at its center if the beam is uniform. Consider the beam to have a mass of 10 kg to help you draw the size of the arrow. Label the force with the symbol Wheam Lastly, draw the upward normal force exerted on the beam by the hinge at the pivot point that serves to counterbalance all of the downward weight forces. Label the force with the symbol n. 6. Calculate Torques: Fill in the table below. Start by listing the downward force exerted on the beam due to the weight of each object (treat g 10 m/s2). Next, record the "lever arm" (distance from pivot) for each object. Next, calculate the magnitude of the torque exerted by each object about the pivot by noting that it is equal to the product of force and lever arm. Lastly, include the appropriate sign for each torque in the table by noting that torques which attempt to rotate bodies counterclockwise are usually treated as positive, while those that try to rotate the system clockwise are treated as negative. Force (N) Lever Arm (m) Torque (N·m) Object Object 1 Object 2 Object 3 0 0 Beam 100

Answer 1

All weights are shown as multiple of g working vertical down

Torque can be calculated using

Torque force x lever armPLEASE upvote the answer and provide feedback

► FBD for up system san Tos L. p-1.75 -1.5 -1.5-- K-0.75 sysg W, we W3 log blog WS um 6 Torgh fore Leverarm Torque (Nom] - 49.05 N -1.5m object 1 - 59 objecte -5 g 7.5g = 93.575 5g = CCW -1.75 85.83 98.1 T=F.d 8.75g 10.09 125g -log 0.75 objects | 5g obijathlesg log -11,25 - -110.36 -CW 1.86 = G ohjects O 0 0 Beam log Lefthand toogue (due to left weigh) E right hand torque (due to right side 257151 Nm = 257.51 Nem mayr)