Thread: Static EQUILIBRIUM of a body

The beams of negligible mass with an overall length of 7m
The beam is supported by a pivot point that is 3m from the left hand side
and 4m from the right

To the right on the pivot, FORCES of 10N and 25 N are applied at a distance of 1m and 4m respectively

to the left of the pivot is an unknown force, F, which is applied 3m from the pivot

the direction of action of each of these forces is as shown

determine the value of the unknown FORCE F, needed to preserved EQUILIBRIUM

is it M=F*d
Moment = Nm
Force = N
distance = Meters

how i work it out?
cheers

Take moments about the frustum point.

F(3) - 10(1)-4(25) = 0

Solve for F

can u explain in more detail plz sorry

Originally Posted by turbod15b

The beams of negligible mass with an overall length of 7m
The beam is supported by a pivot point that is 3m from the left hand side
and 4m from the right

To the right on the pivot, FORCES of 10N and 25 N are applied at a distance of 1m and 4m respectively

to the left of the pivot is an unknown force, F, which is applied 3m from the pivot

the direction of action of each of these forces is as shown

determine the value of the unknown FORCE F, needed to preserved EQUILIBRIUM

is it M=F*d
Moment = Nm
Force = N
distance = Meters

how i work it out?
cheers

I don't know if you in a Math or Physics course. I'll do it ala Physics.

The object is in static equilibrium, meaning that it is neither moving nor rotating. Thus we know two pieces of information: the sum of all forces on the object is 0 N (Newton's 2nd Law) and the sum of all torques on the object about any axis is 0 Nm(rad) (from Newton's 2nd Law applied to rotations.)

We have TWO unknown forces on the beam: F and the unknown upward force exerted by the pivot. All we need is the value of F, so let's use the rotational equilibrium to find it.

The beam does not rotate about ANY axis, so we may choose whatever axis of rotation to calculate the torques as we please. The idea is to pick an axis that helps you out the most, and we can immediately see if we pick the axis where the pivot point is (what would be the physical axis of rotation if the beam were to rotate) then the torque due to the upward force from the pivot is 0 Nm(rad). (Since the "moment arm" of the position of the force to the axis is 0 m.) So the only unknown in calculating the sum of the torques about this axis is F.

I will pick the standard that a counter clock-wise rotation is in the positive sense.

Thus:
Sum(torques) = (3 m)F - (1 m)(10 N) - (4 m)(25 N) = 0
(All the forces are perpendicular to the beam, so we don't have any nasty angles to have to calculate. )


3F - 10 - 100 = 0

3F = 110

F = 110/3 N = 36.67 N

-Dan

thank you