8:27 AM loncapa hep.uprm.edu iPad 79%- Cap12 Points:6 The figure below is a long

Question

8:27 AM loncapa hep.uprm.edu iPad 79%- Cap12 Points:6 The figure below is a long stick with evenly spaced marks. The distance between two adjacent marks is 16.90 cm. For each of the following forces or set of forces determine the requested torque. Use Point F as the axis of rotation. Use the sign convention that a counterclockwise (CCW) torque is considered positive and a clockwise (CW) torque is considered negative. A force of 36.0 N is directed upward perpendicular to the stick at Point C. What is the torque from this force? Submit Answer Tries 4/50 Previous Tries A force of 12.0 N is directed downward perpendicular to the stick at Point G. What is the torque from this force? Submit Answer Tries 6/50 Previous Tries A force of 33.0 N is directed upward perpendicular to the stick at Point E. What is the torque from this force? Submit Answer Tries 0/s A force of 34.0 N is directed upward at angle 36.9° to the right of the perpendicular to the stick at Point G. What is the torque from this force? Submit AnswerTries 0/50 A force of 12.0 N is applied horizontally to the left at G. What is the torque from this force? Submit Answr Tries 0/s0 Two forces are applied simultaneously to the stick. A force of 36.0 N is directed upward perpendicular to the stick at Point C and a force of 33.0 Ns directed upward perpendicular to the stick at Point E. What is the net torque from these two forces? Submit AnswrTries 0/50

Explanation / Answer

Torque is given by

T=rFsin(o)

a)

Here

r=4d=4*0.169=0.676 m

T=36*0.676*sin90

T=24.336 N-m

b)

r=3d =3*0.169=0.507 m

T=0.507*12*sin90

T=6.084 N-m

c)

Here

r=6d =6*0.169=1.014 m

T=1.014*33*sin(-90)

T=-33.462 N-m

d)

Here r=3d=0.507 m

T=0.507*34*sin36.9

T=10.35 N-m

e)

T=0.507*12*sin(0)=0

f)

Net torque

T=(4d)*F*sin90+(6d)*F*sin(-90)

T=0.676*36*1-1.014*33*1

T=-9.126 N-m

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