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Mathematics 7 Online
OpenStudy (anonymous):

If the actual frictional force is equal to 1 lb., then how much effort (force) would be required to drag a 90 lb. box up the plane? 10+ lb. 10- lb. 9+ lb. 9- lb.

OpenStudy (anonymous):

OpenStudy (anonymous):

am i really??

OpenStudy (anonymous):

tbh i guessed im not so sure..

OpenStudy (anonymous):

if friction is 1 lb ,then you are dragging 90+1=91 lb so force required is more than 10 lb. Earlier my conception was wrong.

OpenStudy (anonymous):

\[91 \times \frac{ 1 }{ 9 }=?\]

OpenStudy (anonymous):

so positive 10 is the answer correct?

OpenStudy (anonymous):

or is it negative 10?

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