Problem 11.47 Part A A uniform, 262-N rod that is 2.14 m long carries a 225- N w
ID: 1352204 • Letter: P
Question
Problem 11.47 Part A A uniform, 262-N rod that is 2.14 m long carries a 225- N weight at its right end and an unknown weight W toward the left end (see the figure (Figure 1)). When W is placed 48.8 cm from the left end of the rod, the system just balances horizontally when the fulcrum is located 77.7 cm from the right end. Find W W= 1199.8 Submit My Answers Give Up Incorrect; Try Again; 4 attempts remaining Part B If W is now moved 21.7 cm to the right, how far must the fulcrum be moved to restore balance? Figure 1 of 1 Submit My Answers Give Up Part C 225 N If W is now moved 21.7 cm to the right, in what direction must the fulcrum be moved to restore balance? to the left 0 to the right Submit My Answers Give Up Provide Feedback ContinueExplanation / Answer
forces acting on the rod:
225 N, at the right end
weight 262 N, at 1.07 m from the right end
W, at a distance of 0.488 m from the left end
the fulcrum is at a distance of 0.777 m from the right end
then distance of weight of the rod from the fulcrum=1.07-0.777=0.293 m
distance of W from right end=2.14-0.488=1.652 m
then distance of W from fulcrum=1.652-0.777=0.875 m
hence balancing torque about the fulcrum:
225*0.777=262*0.293+W*0.875
==>W=(225*0.777-262*0.293)/0.875=112.067 N
part B:
let fulcrum is moved to the right by a distance of x m.
then distance of 225 N from fulcrum=(0.777-x) m
distance of weight 262 N from fulcrum=0.293+x m
distance of W from fulcrum=0.875-0.217+x=0.658+x m
then writing torque balancing equation:
W*(0.658+x)+262*(0.293+x)=225*(0.777-x)
==>112.067*(0.658+x)+262*(0.293+x)=225*(0.777-x)
==>x*(112.067+262+225)=225*0.777-262*0.293-112.067*0.658
==>599.067*x=24.3189
==>x=0.0406 m
hence the fulcrum has to be moved 0.0406 m to the right (new distance of flucrum from the right
end is 0.777-0.0406=0.7364 m
part C:
as found out in part B , the fulcrum has to be moved to the right.