20080620

Physics final exam question: cable, boom, and load

Physics 5A Final Exam, spring semester 2008
Cuesta College, San Luis Obispo, CA

Cf. Giambattista/Richardson/Richardson, Physics, 1/e, Problem 8.37

[20 points.] A uniform beam and sign are suspended using a cable that has a breaking strength of 200.0 N, as shown at right. The sign has a weight of 150.0 N and the beam's weight is 60.0 N. The beam's length is 2.00 m and the sign is a square 1.00 m on each side. What is the minimum angle theta allowed between the beam and cable? Show your work and explain your reasoning.

Solution and grading rubric:
  • p = 20/20:
    Correct. In order for the beam and sign to be suspended, the net torque on this system must be zero, and equates this to the sum of the torques of the weight of the beam and sign, and the torque of the cable (using the left end of the beam as the origin). Solves for the minimum angle allowed, which is 45.4 degrees.
  • r = 16/20:
    Nearly correct, but includes minor math errors. May have mismeasured a lever arm.
  • t = 12/20:
    Nearly correct, but approach has conceptual errors, and/or major/compounded math errors. At least has net torque equation set to zero, but problematic calculation of individual torques.
  • v = 8/20:
    Implementation of right ideas, but in an inconsistent, incomplete, or unorganized manner. Some systematic approach at calculating individual torque terms.
  • x = 4/20:
    Implementation of ideas, but credit given for effort rather than merit.
  • y = 2/20:
    Irrelevant discussion/effectively blank.
  • z = 0/20:
    Blank.

Grading distribution:
p: 4 students
r: 7 students
t: 3 students
v: 12 students
x: 2 students
y: 1 student
z: 2 students

A sample of a "p" response (from student 1468) that keeps it short and simple is shown below:
An "r" response (from student 6867) with an error in a lever arm:
A rather flippant "x" response (from student 1337):

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