tendency of rotation in a specified direction. The moment produced by a couple i

Question

tendency of rotation in a specified direction. The moment produced by a couple is called a couple moment. We can determine the moment by finding the sum of the moments of both couple forces about any arbitrary point: PartA A couple (Figure 2) From the figure, Express ViewA so and we can write where r has magnitude d, the perpendicular distance between the forces, and is perpendicular to the forces. Note that, in this equation, all references to point O have been removed. Mc- Submit Figure 1 of 5 (> Part B-Findi The image belo The dimensions What is the net m Express your ans View Available -F

Explanation / Answer

part A:

given :

x=1.4 m

h=0.5 m

F=930 N

total moment is in clockwise direction.

total moment=(4/5)*F*x-(3/5)*F*(h+h)

=483.6 N.m

part B:

given:

F1=650 N

F2=760 N

x1=1.4 m , x2=1.3 m, h=0.15 m, theta=50 degrees, phi=35 degrees

moment due to F1:

net moment in clockwise direction.

net moment=F1*cos(theta)*x1-F1*sin(theta)*h

=510.25 N.m

moment due to F2:

net moment in clockwise direction.

net moment=F2*cos(phi)*x1+F2*sin(phi)*h

=936.97 N

so total moment=510.25+936.97

=1447.2 N.m

part C:

given:

F1=2.5 kN

F2=2.9 kN

F3=3.3 kN

total couple moment on the triangle is required to be 830 N.m counterclockwise.

distance between F3 couple=d*cos(30)

distance between F2 couple=d*sin(30)

distance between F1 couple=d

hence total moment=F2*d*sin(30)+F3*d*cos(30)-F1*d

=1807.9*d

then 830=1807.9*d

==>d=0.4591 m

part D:

given :

F1=3.5 kN

F3=1.3 kN

d=740 mm=0.74 m

net moment required is 290 N.m clockwise

then F1*d-F2*d*sin(30)-F3*d*cos(30)=290

==>F2=(F1*d-F3*d*cos(30)-290)/(d*sin(30))

==>F2=3.9646 kN

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