- #1
letsfailsafe
- 21
- 0
Homework Statement
Questions are included in the photo
Homework Equations
The Attempt at a Solution
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Solve Torque Problems with Expert Help - Homework Question Included"
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SUMMARY
This discussion focuses on solving torque problems related to a beam and a broom, specifically Homework Questions 11 and 12. The key equations involve balancing total clockwise and anti-clockwise torques, with calculations leading to the determination of the beam's weight force (Fb) and the distance needed to balance the broom when an additional weight is added. The correct weight force of the beam is calculated as 17.33N, while the distance required to balance the broom is established as 1.1m. The discussion highlights the importance of accurately identifying pivot points and distances in torque calculations.
PREREQUISITES- Understanding of torque and equilibrium principles
- Familiarity with basic physics equations related to forces and moments
- Knowledge of center of mass concepts
- Ability to perform calculations involving weights and distances
- Study the principles of static equilibrium in physics
- Learn how to calculate torque using the formula τ = r × F
- Explore the concept of center of mass in various shapes and systems
- Practice solving problems involving multiple forces and torques
Students studying physics, particularly those focusing on mechanics, as well as educators looking for examples of torque problems and solutions.
Questions are included in the photo
- #2
Basic_Physics
- 358
- 6
11. There are 3 torques acting on the beam. The sum of these need to be zero.
- #3
aralbrec
- 296
- 1
You should be showing an attempt to a solution before getting help but I'll get you started.
Both questions rely on the same thing: information has been provided to locate the center of mass of the beam (11) and the broom (12).
Both questions rely on the same thing: information has been provided to locate the center of mass of the beam (11) and the broom (12).
- #4
letsfailsafe
- 21
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For question 11.
Total clockwise t = total anti clock wise t
(10 x 3) = (28 x 2) (Fb x ??)
Total tension will be: 28N + 10N + Beam weight force (?)
Problem here is that I don't know the weight force of the beam...
That was my attempt before I post this online...For question 12.
Total support force for a finger must be: (1.5 x 9.8) + (0.4 x 9.8) = 18.62N (?)
I can solve this question if I knew where any two forces were being applied...
Total clockwise t = total anti clock wise t
(10 x 3) = (28 x 2) (Fb x ??)
Total tension will be: 28N + 10N + Beam weight force (?)
Problem here is that I don't know the weight force of the beam...
That was my attempt before I post this online...For question 12.
Total support force for a finger must be: (1.5 x 9.8) + (0.4 x 9.8) = 18.62N (?)
I can solve this question if I knew where any two forces were being applied...
Last edited:
- #5
Basic_Physics
- 358
- 6
Total clockwise t = total anti clock wise t
(10 x 3) = (28 x 2) (Fb x ??)
I assume Fb is the weight of the beam. Its torque will act in the middle of the beam.
(10 x 3) = (28 x 2) (Fb x ??)
I assume Fb is the weight of the beam. Its torque will act in the middle of the beam.
- #6
letsfailsafe
- 21
- 0
Basic_Physics said:
Using point X as the pivot point then:
Total clock wise t = total anti clock wise t
(Fb x 1.5) + (10 x 3) = (28 x 2)
1.5 Fb + 30 = 56
1.5 Fb = 56 -30
Fb = (56-30)/1.5 = 17.3333
17.33N
The answer says 52N...
- #7
Basic_Physics
- 358
- 6
The distance to the middle is 0.5 not 1.5
- #8
letsfailsafe
- 21
- 0
Basic_Physics said:
Oh man... stupid mistake... please help me do the question 12!
- #9
bsbs
- 28
- 0
letsfailsafe said:
total mass 1.5kg
center of mass at 1.4m
imagine the broom is a 2.8m rod instead.
if you add 0.4kg to one side, what is the distance that you need to move from the center, in order to balance the rod.
a logical answer will be <1.4m
- #10
letsfailsafe
- 21
- 0
The answer says its 1.1m...bsbs said:
- #11
Basic_Physics
- 358
- 6
The centre of gravity or centre of mass is located at 1.4 meter. If it was not the broom would have rotated off her finger. She would need to balance it a bit to the right of this point with the additional 400 gram. Call this distance x and recalculate about this new point .
- #12
bsbs
- 28
- 0
letsfailsafe said:
this answer is correct,
next hint,
half of 1.5kg is 0.75 kg
if center is 1.4m, you will need another 1.4m to make this statement true, 2.8m
add 0.4kg to 1.5kg, which make this broom now 1.9kg
with these 3 values, you can easily work the answer as 1.1m
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