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anonymous
 one year ago
two forces P and Q pass through a point A which is 4 ft to the right of and 3 ft above a moment center O. force P is 200 lb directed up to the right at 30degrees with the horizontal and force Q is 100 lb directed up to the left at 60degrees with the horizontal, determine the moment of the resultant of these two forces with respect to O.
anonymous
 one year ago
two forces P and Q pass through a point A which is 4 ft to the right of and 3 ft above a moment center O. force P is 200 lb directed up to the right at 30degrees with the horizontal and force Q is 100 lb directed up to the left at 60degrees with the horizontal, determine the moment of the resultant of these two forces with respect to O.

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rajat97
 one year ago
Best ResponseYou've already chosen the best response.2we have to use \[\tau=F \times r \] first of all we need to break the forces into their componentshorizontal and vertical here we use the formula \[\tau=F _{perpendicular}\times r\] or \[\tau=F \times r_{perpendicular}\] both are the same the formula says that we need to multiply the magnitudes of the force and the radius vector which are perpendicular this is the original condition: dw:1437014598098:dw and when we break the forces into horizontal and vertical components, this becomes the situation: dw:1437014784189:dw we choose the reference point to be the moment center O now we multiply the force components with their perpendicular distances from the moment center. then we determine the direction of the moments due to each component and then add or subtract the moments accordingly and that is the final answer! I may not be very clear so please feel free to ask anything about this

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0how can i determine the sign?im confuse of that...thanks for ur answer by the way..!

rajat97
 one year ago
Best ResponseYou've already chosen the best response.2thanks for the medal! we can determine the sign by using the cross product rule we know, torque is the cross product of radius and the force (the sequence of radius and force is important) you put your fingers in the direction of the radius vector and then curl them towards the force vector and then outstretch your thumb. Then your thumb shows the direction of the moment.

rajat97
 one year ago
Best ResponseYou've already chosen the best response.2well where are you from?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0from philippines..is that the right hand rule u mean?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0my answer is 379.79 lbft

rajat97
 one year ago
Best ResponseYou've already chosen the best response.2yes that's the right hand rule and by the way i am from india

rajat97
 one year ago
Best ResponseYou've already chosen the best response.2i'm working on the answer

rajat97
 one year ago
Best ResponseYou've already chosen the best response.2i got the answer=76.795

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0here my drawing and answer

rajat97
 one year ago
Best ResponseYou've already chosen the best response.2okay! i made the wrong drawing

rajat97
 one year ago
Best ResponseYou've already chosen the best response.2it should be P(y)(4) + Q(y)(4) + Q(x)(3)  P(x)(3)

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0why negative px?is that because of direction of x is negative in P?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0is that mean the answer is positive 379.79lbft

rajat97
 one year ago
Best ResponseYou've already chosen the best response.2i assumed the direction into the plane to be negative and direction and the direction out of the plane to be positive the right hand thumb rule

rajat97
 one year ago
Best ResponseYou've already chosen the best response.2my answer is 376.7949 very close to yours

rajat97
 one year ago
Best ResponseYou've already chosen the best response.2okay then maybe it's correct

rajat97
 one year ago
Best ResponseYou've already chosen the best response.2okay i got it you took into account the direction of the forces and i just manipulated it without taking in account the negative sign of the Qx component

rajat97
 one year ago
Best ResponseYou've already chosen the best response.2the Qx component is directing towards left so it's sign is negative according to the sign convention

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0yes thats what i understand...thats what im asking u..is that correct?is there a general rule for the sign?some says its positive if clockwise and negative when counterclockwise?some says because of direction and u said use a right hand rule...

rajat97
 one year ago
Best ResponseYou've already chosen the best response.2look just forget it! use the right hand rule properly and you'll get the correct answer choose one direction to be positive and the other to be negative generally if not specified in the question, we assume the outward direction to be positive and viceversa. i'll be back in 15 mins. i have to bath

rajat97
 one year ago
Best ResponseYou've already chosen the best response.2thanks for the medal @SurpriseBreeze
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