Can Torque Be Applied to an Object Through Linear Forces Alone?

  • Thread starter Thread starter goldfish9776
  • Start date Start date
  • Tags Tags
    Moment Point
AI Thread Summary
The discussion centers on the application of torque and moments in physics, specifically questioning why a moment isn't calculated as a force multiplied by distance. It clarifies that an applied moment is already defined by its magnitude and direction, and does not require further multiplication by distance. The conversation highlights that torque can be conceptualized as the result of two equal and opposite forces acting along parallel lines. While the theoretical application of moments is acknowledged, the participants note that in practical scenarios, torque is typically generated through linear forces. Ultimately, the discussion emphasizes the distinction between applied moments and the forces that create them.
goldfish9776
Messages
310
Reaction score
1

Homework Statement


Why the moment isn't=15* 3Nm? Since we are taking moment about point O...it's 3m away from O

Homework Equations

The Attempt at a Solution

 

Attachments

  • IMG_20150924_140943.jpg
    IMG_20150924_140943.jpg
    35.8 KB · Views: 424
Physics news on Phys.org
goldfish9776 said:

Homework Statement


Why the moment isn't=15* 3Nm? Since we are taking moment about point O...it's 3m away from O

Homework Equations

The Attempt at a Solution

Are you referring to the 15kNm moment that is applied? You don't multiply that by a distance. (It would give you something with units kNm2.) a force times a perpendicular distance gives a moment, but an applied moment is already a moment. Exactly where it is applied makes no difference, only its magnitude and direction matter.
 
haruspex said:
Are you referring to the 15kNm moment that is applied? You don't multiply that by a distance. (It would give you something with units kNm2.) a force times a perpendicular distance gives a moment, but an applied moment is already a moment. Exactly where it is applied makes no difference, only its magnitude and direction matter.
ya , i knew that . But , how can be moment be applied? only force can be applied , right?
 
goldfish9776 said:
ya , i knew that . But , how can be moment be applied? only force can be applied , right?
Are you asking as a practical matter how it is possible to apply a moment as opposed to a force? There does not need to be a way to do that. Consider turning a nut using a spanner. One can think of it as applying a torque, or as applying two equal and opposite forces along parallel but different lines of action. If you are told a moment of some specified magnitude and direction is applied, you do not need to care about how it is applied.
 
haruspex said:
Are you asking as a practical matter how it is possible to apply a moment as opposed to a force? There does not need to be a way to do that. Consider turning a nut using a spanner. One can think of it as applying a torque, or as applying two equal and opposite forces along parallel but different lines of action. If you are told a moment of some specified magnitude and direction is applied, you do not need to care about how it is applied.
yes, this will only occur in the exercise , but not in daily life ?
 
goldfish9776 said:
yes, this will only occur in the exercise , but not in daily life ?
I cannot think of a way to apply a torque to an object (in an inertial frame) other than by a combination of linear forces.
 
Thread 'Variable mass system : water sprayed into a moving container'

Thread 'Variable mass system : water sprayed into a moving container'

Starting with the mass considerations #m(t)# is mass of water #M_{c}# mass of container and #M(t)# mass of total system $$M(t) = M_{C} + m(t)$$ $$\Rightarrow \frac{dM(t)}{dt} = \frac{dm(t)}{dt}$$ $$P_i = Mv + u \, dm$$ $$P_f = (M + dm)(v + dv)$$ $$\Delta P = M \, dv + (v - u) \, dm$$ $$F = \frac{dP}{dt} = M \frac{dv}{dt} + (v - u) \frac{dm}{dt}$$ $$F = u \frac{dm}{dt} = \rho A u^2$$ from conservation of momentum , the cannon recoils with the same force which it applies. $$\quad \frac{dm}{dt}...
View full post »

Thread 'Struggling to understand the concept of this question (ice cube in water optics question)'

TL;DR Summary: I came across this question from a Sri Lankan A-level textbook. Question - An ice cube with a length of 10 cm is immersed in water at 0 °C. An observer observes the ice cube from the water, and it seems to be 7.75 cm long. If the refractive index of water is 4/3, find the height of the ice cube immersed in the water. I could not understand how the apparent height of the ice cube in the water depends on the height of the ice cube immersed in the water. Does anyone have an...
View full post »

Similar threads

Forces applied to a spring-loaded gas pedal
Replies
7
Views
1K
Three equal forces applied to a rectangle, find net torque direction?
Replies
30
Views
4K
Determining $L_{o}$: Finding Angular Momentum of System
Replies
45
Views
4K
2D rigid body mechanics: force applied on sliding cart
Replies
14
Views
941
How to Apply Derivatives in Physics Problems?
Replies
9
Views
1K
Torque about an accelerating point
Replies
5
Views
3K
Where on a rigid body is a resultant force applied?
Replies
11
Views
1K
Stabilizer Leg Linear Actuator Force to Jack up a Truck's rear tyres
Replies
4
Views
1K
What is the "r" in the moment of inertia?
Replies
13
Views
2K
Book plots position & force on the same graph, still gets right answer
Replies
8
Views
855

Hot Threads

Recent Insights

Back
Top