WEBVTT
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π΄π΅ is a rod having a length of 105 centimetres and negligible weight.
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Forces of magnitudes 240 newtons, 67 newtons, 115 newtons, and 176 newtons are acting on the rod as shown in the figure.
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Given that πΆ and π· are at the trisection of π΄π΅, determine the algebraic sum of the moments of these forces about point π΄.
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Weβre told that the length of the rod is 105 centimetres and that points πΆ and π· are at the trisection.
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105 divided by three is equal to 35.
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Therefore, the distance between π΅ and π·, π· and πΆ, and πΆ and π΄ is 35 centimetres.
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The moment of a force is equal to the force multiplied by the distance from some point.
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In this question, weβre asked to take moments about the point π΄.
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The moment of force π΄ about this point will be equal to 214 multiplied by zero.
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This is equal to zero.
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The moment of force π΅ will be equal to 176 multiplied by 105 as it is 105 centimetres away from π΄.
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This is equal to 18480.
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The force at point πΆ is 67 newtons, and this is 35 centimetres from point π΄.
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67 multiplied by 35 is equal to 2345.
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Finally, the force at π· is 115 newtons.
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This is 70 centimetres away from point π΄.
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115 multiplied by 70 is equal to 8050.
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Weβre told that the forces moving in the anticlockwise direction around point π΄ are positive, and those moving in a clockwise direction are negative.
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Any force moving vertically downwards will move in an anticlockwise direction around point π΄.
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Therefore, the moment at π΅ is positive.
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As the forces at πΆ and π· are acting vertically upwards, these are moving in a clockwise direction.
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Therefore, the moment will be negative.
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The sum of the moments about point π΄ are therefore equal to 18480 minus 2345 minus 8050.
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This is equal to 8085.
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As the forces are measured in newtons and lengths in centimetres, our units will be newton centimetres.
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The correct answer is 8085 newton centimetres.