- #1
Net Force: Identifying and Summing Forces for 2D Surfaces at Angles
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Homework Help Overview
The discussion revolves around identifying and summing forces acting on a brick on a ramp, particularly in a scenario involving a truck and its motion. The subject area includes concepts from dynamics and free-body diagrams in a two-dimensional context.
Discussion Character
- Exploratory, Conceptual clarification, Assumption checking
Approaches and Questions Raised
- Participants discuss the importance of drawing a free-body diagram and selecting an appropriate frame of reference. Questions arise regarding the relationship between the forces acting on the truck and the brick, particularly whether to consider the force parallel to the ramp or the horizontal component of that force.
Discussion Status
Participants are exploring different perspectives on the problem, with some providing hints and clarifications about the forces involved. There is an acknowledgment of the complexity of the situation, particularly regarding inertial versus non-inertial frames. No consensus has been reached, but productive dialogue is ongoing.
Contextual Notes
There are references to attachments that may contain additional information relevant to the problem. The discussion also hints at the need for clarity regarding the orientation of the ramp and the forces acting on the brick.
- #2
Simon Bridge
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You've got your ramp the wrong way around.
Draw a free-body diagram for the brick on the ramp.
hint: pick a frame of reference
(hell: I can't decide on the best hint!)
How you proceed depends on the reference frame - you can do it from the POV of the brick or from the POV of someone standing on the road.
Welcome to PF :)
Draw a free-body diagram for the brick on the ramp.
hint: pick a frame of reference
(hell: I can't decide on the best hint!)
How you proceed depends on the reference frame - you can do it from the POV of the brick or from the POV of someone standing on the road.
Welcome to PF :)
Last edited:
- #3
SammyS
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ten2062 said:
Now we can see it.
- #4
DadOf3
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Simon Bridge said:
The ramps not the wrong way round if the truck travels in reverse or is braking :)
My question, after drawing the free body diagram is should the force of the truck be proportional to the force parallel to the ramp, or the horizontal component of that force since the truck can only accelerate in the horizontal direction?
Thanks.
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- #5
Simon Bridge
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Thrice-kidded: welcome to PF.
Yes, you are correct ... except that the net force on the truck is explicitly to the right in the diagram.
It is just as valid to say that the net force is drawn the wrong way around and more precise to say that the force and ramp orientations are inconsistent.
But less fun :)
Note -
In the inertial frame, the brick is acted on by a net unbalanced force to the right so that it accelerates at the same rate as the truck.
If Fa is the force on the truck and M it's mass then the acceleration of both truck and brick is a=Fa/(M+m).
This means there is a net horizontal force on the brick, to the right, of Ftot=ma: m=mass of the brick.
All the forces in the FBD must sum to this one.
It may be more intuitive to look at it from the brick's perspective - but that involves pseudo-forces since the brick is not in an inertial frame.
The brick is in the same situation as you when you are in a car that is accelerating - it feels a horizontal force to the left pressing it against the ramp.
This and all the other forces on it must sum to zero - since the brick ain't going anywhere inside the truck.
It is usually better to do it the first way - get you used to thinking in terms of inertial frames.
Finesse:
The ramp is at a nice angle:
sin(30)=0.5, cos(30)=0.5√3
Yes, you are correct ... except that the net force on the truck is explicitly to the right in the diagram.
It is just as valid to say that the net force is drawn the wrong way around and more precise to say that the force and ramp orientations are inconsistent.
But less fun :)
Note -
In the inertial frame, the brick is acted on by a net unbalanced force to the right so that it accelerates at the same rate as the truck.
If Fa is the force on the truck and M it's mass then the acceleration of both truck and brick is a=Fa/(M+m).
This means there is a net horizontal force on the brick, to the right, of Ftot=ma: m=mass of the brick.
All the forces in the FBD must sum to this one.
It may be more intuitive to look at it from the brick's perspective - but that involves pseudo-forces since the brick is not in an inertial frame.
The brick is in the same situation as you when you are in a car that is accelerating - it feels a horizontal force to the left pressing it against the ramp.
This and all the other forces on it must sum to zero - since the brick ain't going anywhere inside the truck.
It is usually better to do it the first way - get you used to thinking in terms of inertial frames.
Finesse:
The ramp is at a nice angle:
sin(30)=0.5, cos(30)=0.5√3
Last edited:
- #6
DadOf3
- 2
- 0
Thanks for the welcome Simon, thanks for the appreciation of humour, and thanks for the clarification of the force we need to work with.
Thrice-Thanked.
Now, let's see if Child1 can make sense of the hint, can't just give them the answer can we.
Thrice-Thanked.
Now, let's see if Child1 can make sense of the hint, can't just give them the answer can we.
- #7
Simon Bridge
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Well, if it's three times, it must be true.
Identifying the net force is actually the linchpin for the problem.
Now it's just a matter of identifying the forces you need to sum to get it.
Leaves only ten2062 to get back to us.
Identifying the net force is actually the linchpin for the problem.
Now it's just a matter of identifying the forces you need to sum to get it.
Leaves only ten2062 to get back to us.
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