Alas! you have a flat tyre. Using a jack, you lift the punctured tyre off the...

 

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Alas! you have a flat tyre. Using a jack, you lift the punctured tyre off the ground and attempt to switch the tyre for a spare. You will need to undo the lug-nut that is fastening the wheel onto the axle; to do this you have a standard 4-way wheel brace spanner. For simplicity, you may assume that the wheel is fixed in place (does not rotate about the axle) while you are changing the tyre. Also, you may assume that the wheel is attached by a single lug-nut in the centre, as shown in the diagram: The flat tyre 4-way wheel brace spanner The flat tyre with spanner applied (a) The diameter of your 4-way wheel brace spanner (the distance from the end of the left arm to the end of the right arm) is 0.5 m. You are able to simultaneously exert a force of 100 N downwards with your left hand pushing on the left end of the spanner, and a force of 80N upwards with your right hand pulling on the right end of the spanner. You exert each of these forces perpendicularly to the arms of the spanner. What is the magnitude of the net moment of force that you are able to exert on the lug-mit [2 marks)? (b) The torque you were able to generate in part (a) was not sufficient to loosen the lug-mut. You remember from an engineering class that with greater leverage (a longer moment arm), you are able to generate more torque. You are able to extend your 4-way wheel brace spanner with metal tubes such that it has a diameter of Im. The downside of this is that your arms are no longer long enough to exert forces perpendicular to the spanner, and instead you apply forces at an angle 0 < /2, indicated in the diagram: The flat tyre with wider spanner applied Assume that each force you apply (denoted FL and FR in the diagram) have the same maguitude as in part (a). Under what condition on e will the net torque that you exert on the lug-mut exceed the torque you were able to exert in part (a) [2 marks]? (c) Suppose that you have a mass of 75 kg, and acceleration due to gravity is taken to be 9.8 m/s? downwards. What is the maximum torque you can generate by standing on the left end of the 0.25 m diameter spanner (standing only, no jumping!) [1 mark]? Alas! you have a flat tyre. Using a jack, you lift the punctured tyre off the ground and attempt to switch the tyre for a spare. You will need to undo the lug-nut that is fastening the wheel onto the axle; to do this you have a standard 4-way wheel brace spanner. For simplicity, you may assume that the wheel is fixed in place (does not rotate about the axle) while you are changing the tyre. Also, you may assume that the wheel is attached by a single lug-nut in the centre, as shown in the diagram: The flat tyre 4-way wheel brace spanner The flat tyre with spanner applied (a) The diameter of your 4-way wheel brace spanner (the distance from the end of the left arm to the end of the right arm) is 0.5 m. You are able to simultaneously exert a force of 100 N downwards with your left hand pushing on the left end of the spanner, and a force of 80N upwards with your right hand pulling on the right end of the spanner. You exert each of these forces perpendicularly to the arms of the spanner. What is the magnitude of the net moment of force that you are able to exert on the lug-mit [2 marks)? (b) The torque you were able to generate in part (a) was not sufficient to loosen the lug-mut. You remember from an engineering class that with greater leverage (a longer moment arm), you are able to generate more torque. You are able to extend your 4-way wheel brace spanner with metal tubes such that it has a diameter of Im. The downside of this is that your arms are no longer long enough to exert forces perpendicular to the spanner, and instead you apply forces at an angle 0 < /2, indicated in the diagram: The flat tyre with wider spanner applied Assume that each force you apply (denoted FL and FR in the diagram) have the same maguitude as in part (a). Under what condition on e will the net torque that you exert on the lug-mut exceed the torque you were able to exert in part (a) [2 marks]? (c) Suppose that you have a mass of 75 kg, and acceleration due to gravity is taken to be 9.8 m/s? downwards. What is the maximum torque you can generate by standing on the left end of the 0.25 m diameter spanner (standing only, no jumping!) [1 mark]?

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Required solution First we need to draw the FBD free body diagram to accomplish the ... View the full answer