Calculating Acceleration of a Hay Bale Being Lifted Onto a Barn Loft

  • Thread starter Thread starter jen333
  • Start date Start date
  • Tags Tags
    Dynamics
Click For Summary
SUMMARY

The discussion revolves around calculating the acceleration of a hay bale being lifted using a tension force of 850N and a mass of 81.6kg. The correct acceleration, derived from the net force after accounting for gravitational force, is approximately 0.607m/s². The user initially calculated the gravitational force as 800N, leading to a slight discrepancy in their final answer. The accurate calculation reveals that the gravitational force is 800.496N, resulting in a net force of 49.504N and an acceleration of 0.6067m/s².

PREREQUISITES
  • Understanding of Newton's Second Law (F=ma)
  • Ability to draw and interpret Free Body Diagrams (FBD)
  • Knowledge of gravitational force calculations
  • Familiarity with basic physics concepts related to dynamics
NEXT STEPS
  • Study advanced applications of Newton's laws in dynamic systems
  • Learn about Free Body Diagram techniques for complex systems
  • Explore the effects of rounding in physics calculations
  • Investigate tension forces in various lifting scenarios
USEFUL FOR

This discussion is beneficial for physics students, educators, and anyone interested in understanding dynamics and force calculations in real-world applications.

jen333
Messages
58
Reaction score
0
Hey!
I have a physics problem on dynamics. I did the question and received an answer, however, the answer is different from the one that i was given. I just want to know if I did something wrong (i probably did) or if the answer that was given is wrong. here it goes:

Proper storage of hay bales is essential for the survival of cattle during Canadian winters. A farmer uses a cable to exert an upward pull of 850N on a 81.6kg bale of hay at a local farm. Use a diagram and determine the acceleration of the bale as it is lifted onto a barn loft.
The answer that is given is 0.607m/s^2

Here's how I executed the problem:
a) drew a FBD diagram
^
l
l
l T=850N
O
l
\/ Fg

b) i found the force of support on the bale of hay (not sure if this is right)
F=ma
= 81.6kg x 9.81m/s^2
= 800N

c) subtracted this force from the tension force:
850N-800N= 50N

d) i used the formula again F=ma

a=F/m= 50N/81.6kg
= 0.613m/s^2
 
Physics news on Phys.org
Considering you are off by 0.06m/s^2 I would say your error is from rounding.

In part b) 81.6 x 9.81 = 800.496

850 - 800.496 = 49.504

49.504/81.6 = 0.606666666m/s^2
 
alright! good to know that I'm on the right track :biggrin:
Thx for the help!

jen
 

Similar threads

  • · Replies 30 ·
2
Replies
30
Views
4K
Replies
7
Views
2K
  • · Replies 9 ·
Replies
9
Views
3K
  • · Replies 9 ·
Replies
9
Views
5K
  • · Replies 15 ·
Replies
15
Views
2K
  • · Replies 4 ·
Replies
4
Views
2K
  • · Replies 3 ·
Replies
3
Views
3K
  • · Replies 2 ·
Replies
2
Views
5K
  • · Replies 15 ·
Replies
15
Views
5K
Replies
9
Views
4K