Friction causes a torque - then why do we sum it as a regular force?

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SUMMARY

Friction generates torque, which is not a force and therefore is not included in Free Body Diagrams (FBDs). Torque must be balanced by another torque, not a force, as demonstrated in rigid body dynamics. The net force is calculated using F=ma, while net torque is calculated using τnet=Iα. Understanding the distinction between forces and torques is crucial for analyzing systems with both translational and rotational motion.

PREREQUISITES
  • Rigid Body Dynamics
  • Free Body Diagrams (FBD)
  • Newton's Second Law (F=ma)
  • Torque and Moment of Inertia (τnet=Iα)
NEXT STEPS
  • Study the principles of Rigid Body Dynamics in detail.
  • Learn how to construct and analyze Free Body Diagrams (FBDs).
  • Explore the relationship between torque and angular acceleration.
  • Investigate static equilibrium conditions in rigid body systems.
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Physics students, mechanical engineers, and anyone studying dynamics involving both translational and rotational motion.

Differentiate it
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Homework Statement
Friction causes a torque which is balanced out by the normal force. In most physics problems, we simply sum the applied force and friction, without considering anything about torque. I'm kind of new to Rigid Body Dynamics, so it'd be helpful if someone also explained how translational forces and torque are added in a system where there is both translational and rotational motion and/or sent links to some online resources. Thanks!
Relevant Equations
F=ma
Screenshot_20220829-112957.jpg
 
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Differentiate it said:
Homework Statement:: Friction causes a torque which is balanced out by the normal force. In most physics problems, we simply sum the applied force and friction, without considering anything about torque. I'm kind of new to Rigid Body Dynamics, so it'd be helpful if someone also explained how translational forces and torque are added in a system where there is both translational and rotational motion and/or sent links to some online resources. Thanks!
Relevant Equations:: F=ma

View attachment 313399
Torque is not a force so we do not include it on a FBD. We also don't sum it along with forces. (Check the units... the units for torque and force are different.)

I don't know what you mean by "Friction causes a torque which is balanced out by the normal force." A torque can only be balanced by another torque, not a force, and the torque due to the friction force is not canceled by the torque due to a normal force.

-Dan
 
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topsquark said:
Torque is not a force so we do not include it on a FBD. We also don't sum it along with forces. (Check the units... the units for torque and force are different.)

I don't know what you mean by "Friction causes a torque which is balanced out by the normal force." A torque can only be balanced by another torque, not a force, and the torque due to the friction force is not canceled by the torque due to a normal force.

-Dan
Right right, I'm just stupid, no probs no probs i get it
 
Differentiate it said:
Friction causes a torque which is balanced out by the normal force.
The torque caused by a force is always in respect of some chosen axis. In the diagram, using an axis at ground level, F applies a torque, Ff doesn’t.
The exception is when you have a pair of equal and opposite forces along parallel lines of application. In that case the net torque is the same no matter what axis you choose.
What you can say is that if the system is static then the torque due to the F, Ff pair balances the torque due to the N, mg pair.
Differentiate it said:
how translational forces and torque are added in a system
You add the forces to write Fnet=ma, and add the torques to write τnet=Iα.
topsquark said:
Torque is not a force so we do not include it on a FBD.
It's quite standard to include torques in an FBD where appropriate. In some posed questions, applied torques are specified as such, rather than as pairs of equal and opposite forces.
 
Differentiate it said:
Right right, I'm just stupid, no probs no probs i get it
If you were stupid you wouldn't have asked us to help you clarify your question!

-Dan
 
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Likes gmax137, hmmm27, Lnewqban and 3 others
The book claims the answer is that all the magnitudes are the same because "the gravitational force on the penguin is the same". I'm having trouble understanding this. I thought the buoyant force was equal to the weight of the fluid displaced. Weight depends on mass which depends on density. Therefore, due to the differing densities the buoyant force will be different in each case? Is this incorrect?

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