- #1
letsfailsafe
- 21
- 0
Homework Statement
Questions are included in the photo
Homework Equations
The Attempt at a Solution
Last edited:
Solve Torque Problems with Expert Help - Homework Question Included"
- Thread starter letsfailsafe
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- Torque
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Homework Help Overview
The discussion revolves around torque problems involving a beam and a broom, focusing on the application of torque equilibrium principles. Participants are examining the conditions under which the sum of torques is zero and how to locate the center of mass for the objects involved.
Discussion Character
- Exploratory, Assumption checking, Problem interpretation
Approaches and Questions Raised
- Participants discuss the need to balance torques acting on the beam and the broom, with some attempting calculations based on provided weights and distances. Questions arise regarding the weight force of the beam and the correct distances to use for torque calculations.
Discussion Status
There are multiple interpretations of the torque calculations, with participants providing their reasoning and calculations. Some guidance is offered regarding the center of mass and the implications of adding weight to one side of the broom. However, there is no explicit consensus on the correct approach or final answers.
Contextual Notes
Participants are working under the constraints of homework rules, which may limit the amount of direct help they can receive. There is also uncertainty regarding the values of certain forces and distances, which affects the calculations being discussed.
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
- 0
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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