Naked Science Forum
Non Life Sciences => Technology => Topic started by: James on 30/06/2016 20:57:05
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I am currently studying for a biomechanics test next week. I've been given two weeks notice, and with no help from tutors (as they are all away on holiday [>:(] ) I have been left to my own devices.
I am not looking for the answer, only how to go about getting the answer for the attached question.
Sorry if this is trivial to most [:P]
Thanks in advance [:)]
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You need to look up levers and torque.
I assume you have been given simpler examples previously as guidance.
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Thanks Colin.
I have been given examples which are somewhat a breakdown of this question, however I was unsure as to how to go about doing the latter part of the question where by the Fpectoralis comes in to play.
Would this be a case of working out the vector product for the upper arm muscle forces and incorporate this into the former part of the question? Or would I have to work backwards from the first part of the question?
Thanks for your help.
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... I was unsure as to how to go about doing the latter part of the question where by the Fpectoralis comes in to play.
Would this be a case of working out the vector product for the upper arm muscle forces
That's the way I would read it, so the horizontal vector component of the Fpectoralis balances the torque from the forearm
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One would have to study the bone structure where the muscle is attached from memory I believe about ten time the force is needed by the bicep to lift a given weight in the hand
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One would have to study the bone structure where the muscle is attached from memory I believe about ten time the force is needed by the bicep to lift a given weight in the hand
In this case all the details of the attachment points are given in the diagram
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I am currently studying for a biomechanics test next week. I've been given two weeks notice, and with no help from tutors (as they are all away on holiday [>:(] ) I have been left to my own devices.
I am not looking for the answer, only how to go about getting the answer for the attached question.
Sorry if this is trivial to most [:P]
Thanks in advance [:)]
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For answering question a) you use the elbow as the axis of rotation, and assume that the upper arm is rigid so you can ignore torque components there.
For answering question b) you use the top of upper arm as the axis of rotation, and assume that the elbow rigidly attach lower arm to the upper arm. So you need to calculate all torque components there with top of the upper arm as reference to calculate the distance and angle of each force shown in the diagram.
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The force required is best viewed by looking at the point of pivot of the shoulder.
Work out all the various moments about that point that the various loads will produce. Add them all up and then work out the little similtaneous equasion where the vertical lift is made up of the top force and the pectoral force x sin 30 .
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The force required is best viewed by looking at the point of pivot of the shoulder.
No it isn't.
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The force required is best viewed by looking at the point of pivot of the shoulder.
No it isn't.
the force required by the shoulder muscles is best.....
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For answering question a) you use the elbow as the axis of rotation, and assume that the upper arm is rigid so you can ignore torque components there.
For answering question b) you use the top of upper arm as the axis of rotation, and assume that the elbow rigidly attach lower arm to the upper arm. So you need to calculate all torque components there with top of the upper arm as reference to calculate the distance and angle of each force shown in the diagram.
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Thank you for your reply.
When I am dealing with question (b), should i negate the answer to part (a) or carry it through? It seems like there will be no rotation at the shoulder axis as the forces that keep it it in equilibrium will be the same as part (a) as they are acting parallel to the Y-axis?
The only additional information I think that will come in to play will be the force acting downward from the weight of the upper arm, which will have no rotation movement as it acts through the Y-axis.
I'm sorry if I seem confused by this, I have this module as part of my biomedical engineering course and sadly my lecturers assume everyone in my class has the same knowledge base. I don't even have the answers to check my workings out.
Thank you
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When I am dealing with question (b), should i negate the answer to part (a) or carry it through?
I'm not sure what you mean by this.
It seems like there will be no rotation at the shoulder axis as the forces that keep it it in equilibrium will be the same as part (a) as they are acting parallel to the Y-axis?
That's right except you need to add the effect of the horizontal component of the Fpectoralis - lever arm and force. But, obviously, don't add the Fbiceps in [;)]
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When I am dealing with question (b), should i negate the answer to part (a) or carry it through?
I'm not sure what you mean by this.
It seems like there will be no rotation at the shoulder axis as the forces that keep it it in equilibrium will be the same as part (a) as they are acting parallel to the Y-axis?
That's right except you need to add the effect of the horizontal component of the Fpectoralis - lever arm and force. But, obviously, don't add the Fbiceps in [;)]
Thanks Colin.
What I meant to ask was if the Fbicep in question (a) is used to balance the body and bring about a state of equilibrium, the only component that would be different in question (b) would be the downward force from the upper arm weight.
Therefore the force required to counteract this would be the Fdeltoid, and perhaps a Y-axis vector component that may make up the Fpectoralis also. However my concern with this would be the delegation of this force between the Fdeltoid and Fpectoralis muscles.
To me, question (b) seems pointless, or somewhat too basic to be worth 15points of a 25point question.
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since the pivot point in a) is different than b) you need to recalculate each torque components, which are cross product of force vector and distance vector, and then sum them all up.
point b worth more point because it have more torque components than point a hence you have to do more calculations.
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What I meant to ask was if the Fbicep in question (a) is used to balance the body and bring about a state of equilibrium, the only component that would be different in question (b) would be the downward force from the upper arm weight.
I misunderstood what assumptions you were making.
As Hamdani Yusuf says, the pivot points are different.
In a) the bicep holds the forearm in equilibrium around the elbow as a pivot.
In b) the Pectoralis holds equilibrium around the shoulder as pivot, and the deltoid resists the dislocating force of the weight in the hand, the forearm weight, and - don't forget - the mass of the upper arm, which wasn't needed in the calculation in a). This is a very different calculation to a).