Is there an error in the answer for Impulse Momentum Linear Homework?
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Homework Help Overview
The discussion revolves around a problem related to impulse and momentum in a linear dynamics context, specifically focusing on parts c and d of a homework question involving forces acting on a hammer and pile system. Participants are checking the correctness of their answers and exploring the implications of various forces, including gravity and resistive forces.
Discussion Character
- Mixed
Approaches and Questions Raised
- Participants are attempting to verify their calculations for parts c and d, questioning the application of forces such as gravity and resistive forces. There is discussion about the net force and its role in determining the motion of the system, as well as confusion regarding the signs of forces in their equations.
Discussion Status
Some participants have provided insights into the forces acting on the hammer and pile, while others express confusion about the relationships between these forces and the net force. There is ongoing exploration of how to set up equations to solve for the resistive force and clarify the role of gravity in the context of the problem.
Contextual Notes
Participants are grappling with the definitions and roles of various forces, including the gravitational force and the net force, as well as how these relate to the impulse-momentum principle. There is mention of specific values and conditions, such as the mass of the hammer and the external forces involved, which may influence the calculations.
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I don't get anywhere near their answers for (a) or (b).freshbox said:yes but not for c and d
I get their answers only if I change the force to 120 MN.
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ehild
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NascentOxygen said:I don't get anywhere near their answers for (a) or (b).
I get their answers only if I change the force to 120 MN.
Have you used 8000 kg as mass of the hammer? (Mg = 103 kg)
ehild
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Ah, with that correction all of my answers now are the same as those given alongside the question.ehild said:Have you used 8000 kg as mass of the hammer? (Mg = 103 kg)
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M1=mv
=(9000)(8.82)
=79380
M2=mv
=0
I1-2=Force x Time
=(F)(0.1)
=0.1F
M1+I1-2=M2
79380-0.1F=0
79380=0.1F
F=793800 Ans Wrong
Counting gravity force into working
I1-2= Force x Time
=(9.81F)(0.1)
=0.981F
79380=0.981F
F=80917.43 Ans Wrong
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That's the net force acting. Use that to find the resistive force.freshbox said:Part C
M1=mv
=(9000)(8.82)
=79380
M2=mv
=0
I1-2=Force x Time
=(F)(0.1)
=0.1F
M1+I1-2=M2
79380-0.1F=0
79380=0.1F
F=793800 Ans Wrong
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I'd say so.freshbox said:Is I1-2= Force x Time this force always the net force?
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Draw a free body diagram and analyze the forces acting on the hammer+pile system.freshbox said:Can you guide me on how to use the net force to find the resistive force? Is there a formula to it?
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For one thing, action and reaction forces act on different bodies. What you want are the forces acting on the hammer+pile.freshbox said:There is an action and reaction force. Action :+793800 Reaction-793800![]()
Second, that 793800 N force is a net force, not an individual force. So even talking about it as an "action" is meaningless.
What forces act on the hammer+pile? Hint: Only two forces act.
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That's the opposing force to bring about the deceleration. In addition, you also have to oppose their weights, as well.freshbox said:Part C
M1=mv
=(9000)(8.82)
=79380
M2=mv
=0
I1-2=Force x Time
=(F)(0.1)
=0.1F
M1+I1-2=M2
79380-0.1F=0
79380=0.1F
F=793800 Ans Wrong
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What about the force you are asked to find?freshbox said:I just can't think of any other forces besides gravity..
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How many Newtons force UP must be exerted to balance their weights acting DOWNWARD? -793800N? What are you trying to tell me, action/reaction ?
Urgggh... so confusing I don't know how to use the net force to solve :(
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So, what's the gravitational force? Set the sum equal to the net force.freshbox said:Yes the 2 forces are gravity and resistive force, sorry.
Forget about 'balancing' anything.How many Newtons force UP must be exerted to balance their weights acting DOWNWARD? -793800N? What are you trying to tell me, action/reaction ?
There are two forces, one unknown. And you already know the net force, so set up an equation to solve for the resistive force.
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Fr=793800+88290
Fr= 882090
Ok this is the answer but I still don't understand the concept behind it.
Doc AL said, So, what's the gravitational force? Set the sum equal to the net force. I assume G Force = Net Force -> can you explain why G force is equal to net force?
From this equation F=Fr-G, how do you determine G force is -ve, Fr and F is both +ve because at first I tried setting it as G=F-Fr because I thought all should be positive except Fr since it is "resist"
Please explain, thank you guys so much
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ehild
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OK.freshbox said:793800=Fr-88290
Fr=793800+88290
Fr= 882090
The "G force" is just the weight of the system, which acts down. It's one of the two forces acting. It's not the net force.Ok this is the answer but I still don't understand the concept behind it.
Doc AL said, So, what's the gravitational force? Set the sum equal to the net force. I assume G Force = Net Force -> can you explain why G force is equal to net force?
The resistive force is upward (the ground pushes up), the gravitational force is downward; the net force is upward. So, taking up as positive, the gravitational force would be negative.From this equation F=Fr-G, how do you determine G force is -ve, Fr and F is both +ve because at first I tried setting it as G=F-Fr because I thought all should be positive except Fr since it is "resist"
What you did above, which is correct, is to set the net force equal to the sum of the individual forces:
ƩF = [Resistive Force] + [Gravity] (think of this as a vector equation)
Using up as positive:
793800 = Fr - mg
So Fr = 793800 + mg
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ehild
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