A pivot is placed at Point K and a force of 15.0 N is directed downward perpendi

Question

A pivot is placed at Point K and a force of 15.0 N is directed downward perpendicular to the stick at Point J. What is the magnitude and direction of the force which must be applied at point B in order to obtain static equilibrium? Upward or Downward

A pivot is placed at Point D and a force of 10.0 N is directed downward perpendicular to the stick at Point H. One at a time, each of the following 3 forces are applied to the stick in the manner described. Where must these forces be applied in order to achieve static equilibrium? If the is no point that will work within reason, select the option None.

A force of 7.5 N directed upward perpendicular to the stick.

A force of 7.5 N directed downward perpendicular to the stick.

A force of 2.7 N directed upward perpendicular to the stick.

Explanation / Answer

given, length of the stick, l

and d = 12.5 cm

a. A pivot is placed at A

F = -19 N at J ( downwards)

force applied at F for static equilibrium = f

from moemnt balance about A

f*5d = F*4d

f = 4F/5 = -15.2 N ( downwards)

b. pivot at point K

F = -15 N at point J

force appliet at B = f

then from moment balance about A

F*8d + f*9d = 0

f = -8F/9 = 13.333 N ( upwards)

c. pivot at point D

F = -10 N at point H

force f = +7.5 N

then form moment balance

F*3d + f*x = 0

x = -3Fd/f = -3d(F/f) = 4d

hence a force must be applied at E

f = -7.5 N

from moment balance

F*3d = f*x

x = 3d(F/f) = 4

hence f must be applied at point F

f = 2.7 N

form moment balance

F*3d + f*x = 0

x = -3d(F/f) = 11.11d

hence this foirce cannot be applied on the scale

Related Questions

Navigate

• Browse (All)
• Browse A
• Subjects

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.