ATAR Notes: Forum
VCE Stuff => VCE Mathematics => VCE Mathematics/Science/Technology => VCE Subjects + Help => VCE Specialist Mathematics => Topic started by: wildareal on October 07, 2010, 09:37:18 pm
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If a boat is being pulled by two forces, of 40 kg wt towards the east and 30 kg wt to the
north-west, what third force must be acting on the boat if it remains stationary? Give the
magnitude and direction. Thanks.
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How bad is statics!
Draw out the diagram, and then use sine/cosine rule to find angle/length
Hope that helps
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How bad is statics!
Draw out the diagram, and then use sine/cosine rule to find angle/length
Hope that helps
Haha yer! I think with this question you draw a triangle with side T1, 40kgwt and angle 45* and solve, since the unknown is gravity as the two other forces NW and E.
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take the i direction (east): 40 - 30cos(45) + F = 0
so F= (15(sqrt(2))-40)i
Take the j direction (north): 30sin45 + F=
so F=-15sqrt(2)j
add them together and you get F=(15(sqrt(2))-40)i - 15sqrt(2)j
then simply use tank(y/x) for the angle, and the magnitude is easy peasy yr 10 math
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take the i direction (east): 40 - 30cos(45) + F = 0
so F= (15(sqrt(2))-40)i
Take the j direction (north): 30sin45 + F=
so F=-15sqrt(2)j
add them together and you get F=(15(sqrt(2))-40)i - 15sqrt(2)j
That looks good. Did you make East the i axis and N the j axis?
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take the i direction (east): 40 - 30cos(45) + F = 0
so F= (15(sqrt(2))-40)i
Take the j direction (north): 30sin45 + F=
so F=-15sqrt(2)j
add them together and you get F=(15(sqrt(2))-40)i - 15sqrt(2)j
That looks good. Did you make East the i axis and N the j axis?
You sure did! Thanks!
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How bad is statics!
Draw out the diagram, and then use sine/cosine rule to find angle/length
Hope that helps
Haha yer! I think with this question you draw a triangle with side T1, 40kgwt and angle 45* and solve, since the unknown is gravity as the two other forces NW and E.
btw, gravity isn't involved - the boat isn't in free fall - it's being pulled perpendicular to the earth's surface
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Yes that makes sense. Does that mean there is a normal force on the water?
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Was the attached the right way to do it if I remember correctly?
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Yes that makes sense. Does that mean there is a normal force on the water?
Yes there is a normal force, (*and strictly speaking, buoyancy effects and water drag should be taken into account*), but you can disregard all upward/downward motion completely for the purposes of this question (and any similar questions in the VCE spec course). Just deal with a point on a flat Cartesian plain.