Vector component of the weight of an object

Click For Summary
SUMMARY

The discussion focuses on calculating the weight of an object that exceeds the capacity of a scale by using a support system. The method involves resting one edge of the object on the scale while the other edge is supported at a higher point, creating an angle θ with the horizontal. The force exerted on the scale is determined by the equation F = cos(θ) x (1/2)W, where W is the total weight of the object. This approach is particularly applicable to uniform objects, such as a length of sawn timber.

PREREQUISITES
  • Understanding of basic physics concepts, specifically forces and angles
  • Familiarity with trigonometric functions, particularly cosine
  • Knowledge of static equilibrium and support forces
  • Experience with practical applications of weight measurement techniques
NEXT STEPS
  • Research the principles of static equilibrium in physics
  • Learn about trigonometric functions and their applications in force calculations
  • Explore practical methods for measuring weights beyond scale capacities
  • Investigate the use of supports and levers in weight distribution scenarios
USEFUL FOR

Physics students, engineers, and anyone interested in practical applications of weight measurement and force calculations in static systems.

Philpense
Messages
2
Reaction score
0
Graduated College over thirty years ago. In physics, I recall a a question where one is required to weigh an object beyond the capacity of the scale. The answer involved resting one edge of the object on scale and the other on a separate support higher than the scale such that the object created and angle from the perpendicular . I believe that the cosine of the angle from the perpendicular played in the solution. Guidance sought
 
Physics news on Phys.org
Hi Philpense. http://img96.imageshack.us/img96/5725/red5e5etimes5e5e45e5e25.gif

It sounds do-able. Consider something uniform like a length of sawn timber. Providing you can ensure the upper support prevents the beam slipping, then it appears the force on the lower support (i.e., the scales) will be cos θ x half the weight[/color], where θ is the beam's inclination to the horizontal.
 
Last edited by a moderator:

Similar threads

  • · Replies 6 ·
Replies
6
Views
2K
  • · Replies 13 ·
Replies
13
Views
4K
  • · Replies 9 ·
Replies
9
Views
2K
  • · Replies 8 ·
Replies
8
Views
3K
  • · Replies 5 ·
Replies
5
Views
6K
Replies
3
Views
10K
  • · Replies 3 ·
Replies
3
Views
4K
  • · Replies 5 ·
Replies
5
Views
3K
  • · Replies 6 ·
Replies
6
Views
2K
Replies
9
Views
13K