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UnsolvedEqn
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For my maths homework I'm stuck on this question. Could someone work this out and tell me the method maybe. Thanks.
A loaded skip of mass 2000kg rests on a path inclined at 21 degrees to the horizontal. A force of magnitude 2500N acts upwards on a skip in a direction parallel to a line of greatest slope of the path
1. Find normal force by the path
2. Find frictional force exerted on the skip.
A loaded skip of mass 2000kg rests on a path inclined at 21 degrees to the horizontal. A force of magnitude 2500N acts upwards on a skip in a direction parallel to a line of greatest slope of the path
1. Find normal force by the path
2. Find frictional force exerted on the skip.
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theone
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Draw a diagram, write in the forces you know, mark on the friction and the resultant forces and solve.
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Rysh
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i dont think u typed it 100% accurately but i think i understand what they want.
a) u want the normal force exerted by the path. the normal is perpendicualr to the slope. remember always label your forces and make all forces leave the object, its much easier. anyway heres the method.
mgcos21
which means ( 2000 x 9.8 ) x cos21 that should give u the answer
b) normally it should go down the plane but in this case it will go up. friction always opposes the motion. now look at ur forces u will see if u were to imagine all forces parallel across the slope, 1would be 2500N force the other would be the friction force which u should have drawn PLUS the meight of the skipXsin21. u may know this but just to be sure the normal weight doenst count coz its value is Ycos90. cos90 is 0!.
so anyway the equation is : F + (2000x9.8) = 2500
re-arrange and solve and thats ur answer.
later
a) u want the normal force exerted by the path. the normal is perpendicualr to the slope. remember always label your forces and make all forces leave the object, its much easier. anyway heres the method.
mgcos21
which means ( 2000 x 9.8 ) x cos21 that should give u the answer
b) normally it should go down the plane but in this case it will go up. friction always opposes the motion. now look at ur forces u will see if u were to imagine all forces parallel across the slope, 1would be 2500N force the other would be the friction force which u should have drawn PLUS the meight of the skipXsin21. u may know this but just to be sure the normal weight doenst count coz its value is Ycos90. cos90 is 0!.
so anyway the equation is : F + (2000x9.8) = 2500
re-arrange and solve and thats ur answer.
later
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