Torque AP1 Assessment Journal #co Tuesday 2/27/2018 The diagram below shows a me
ID: 3308925 • Letter: T
Question
Torque AP1 Assessment Journal #co Tuesday 2/27/2018 The diagram below shows a meterstick restingion two ring stands. Different masses are hung from the meterstick, which has amasof- 135grams. A mass m1=250g is suspended from the meterstick at the 20.0 cm markd a mass of ma= 500gis suspended from the 30.0 cml mark and a mass ofms=150lis suspended from the 700enf- mark of the meterstick as shown. What will be the magnitude and point of applicatiom of a single upward force that can lift this _meterstick while maintaining the horizontal position. m2 m, a.) Complete a force diagram & select a center of rotation! That starts out as 0,) what is the magnitude of the upward force? ) What is the point of application of this single upward force ON THE OPPOSITE FACING PAGE WRITE Answers & Explanations .Enduring understanding - . Essential knowledge Science PracticesExplanation / Answer
The single upward force required to lift the entire system will be:
W = (M+m1+m2+m2)g = (135 + 250 + 500 + 150) x 10-3 x 9.8 = 10.143 N
its point of application should be chosen such that the net torque should be zero (sum of clockwise moments = sum of anticlockwise moments).
Since the ruler should be horizontal, the normal force from both the contacts should be equal
so, 2R = W = 10.143
=> R = 5.0715 N
now, taking moments about the left point of contact and assuming clockwise moments to be negative,
M = 5.0715(0) - 0.25(9.8)(0.2) - 0.5(9.8)(0.4) - 0.135(9.8)(0.5) - 0.15(9.8)(0.7) + Wd = 0
=> 5.0715(0) - 0.25(9.8)(0.2) - 0.5(9.8)(0.4) - 0.135(9.8)(0.5) - 0.15(9.8)(0.7) + 10.143d = 0
=> d = 0.408 m = 40.8 cm
so, the single upwards force should be of magnitude 10.143 N and it should applied at 40.8 cm from the left for equilibrium.