Engineering Statics - Moments problem

In summary, the conversation discusses a question about computing the moment of a force about a specific axis using the right hand rule and resolving forces. The conversation also addresses a mistake in the original question and clarifies when to use the BA vector and when to use the AB vector. It also briefly discusses the use of longitudinal forces in relation to strings, cables, and other similar objects.
  • #1
yugeci
61
0

Homework Statement



I have attached the diagram. The question says "Compute the Moment Mo of the 1.2kN force about the axis O-O.

Homework Equations



M = r x F
AB = OB - OA
Resolving forces, Fx = F cos Θ.. Fy = F sin Θ etc.

The Attempt at a Solution



Here are the axes I've made, using the Right Hand Rule:

jPQSfyR.png


OA = (-300, 0, 105)
OB = (0, 0, 200)

So AB = (300, 0, 95) .. this is the r vector.

For the F vector, Fz = 1.2 sin 60 = 1.04 kN
Fxy = 1.2 cos 60 = 0.6 kN

Fx = -0.6 cos 50 = -0.39 kN
Fy = 0.6 sin 50 = 0.46 kN

M = r x F (since we need the moment about the O-O axis, I'll just get the j component).

| 300 0 95 |
| -0.39 0.46 1.04 |

Solving gives M @ o-o as -349.05 Nm. However the answer is +327 Nm. What have I done wrong?
 

Attachments

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  • #2
I am missing why the y-component of the r-vector should be zero...
[Edit] sorry, axes mixed up. Will re-do.
 
Last edited:
  • #3
Turns out you also have something mixed up (means we're even...):

The r-vector is BA, not AB, so (-300, 0, -95)​

Doesn't help us to find a different value, though. Checked a few times.

So have some confidence in your result (and tell me what we did wrong once you find out :smile:).
 
  • #4
It turns out that we are right, I think. The original question (older edition) was in lb-in, and I guess there was some error when the values got converted to N-mm. You are right that we need the BA vector instead of AB.. but that's just a sign change.. which leads me to my next question?

When do we use AB, and when do we use BA (the reverse)? Here are two example questions:

awJi4GD.png


In the first we use BA. In the second we use AB.

Thanks in advance.
 
  • #5
Help? :\
 
  • #6
The word "on" is key here!

Antenna case:
On point B a force pulls in the direction BA. (So that force will have a positive z-component, positive y, negative x). But on point A the cable to B exercises a force that is in the direction AB (so all components opposite sign).
Can't say i like the picture very much, because the y-coordinate of B is pretty vague.
Does the exercise continue with something that involves assuming that on point A the sum of forces has no x- and y-components?

Turnbuckle case :
On point A which is part of member AD a force pulls in the direction AB. (So that force will have a negative z-component, positive x, positive y)

Help? :\
was out of town & off line in the weekend. Happens ;-)

PS Generally: nice pictures!

PS strings, rubber bands, springs, ropes, cables: in simple cases they are supposed to exercise longitudinal (in line, as opposed to transversal, perpendicular to..) forces (otherwise the ends would move sideways).
The forces on a string, cable, section of string, etc. point outwards, the forces the string, cable, etc. exercises on the stuff attached point "inwards". If the rope itself is weightless AND is not accelerated, all are equal in magnitude.
 

1. What is the definition of a moment in engineering statics?

A moment in engineering statics is a measure of the tendency of a force to cause a body to rotate about a specific point or axis. It is calculated by multiplying the magnitude of the force by the perpendicular distance from the point or axis to the line of action of the force.

2. How do you calculate the magnitude and direction of a moment?

The magnitude of a moment can be calculated by multiplying the magnitude of the force by the perpendicular distance from the point or axis to the line of action of the force. The direction of the moment is determined by the right-hand rule, where the direction of the moment is perpendicular to the plane formed by the force and the perpendicular distance.

3. What is the principle of moments in engineering statics?

The principle of moments states that for a body in equilibrium, the sum of the clockwise moments about any point is equal to the sum of the counterclockwise moments about the same point. This principle is used to solve moments problems in engineering statics.

4. How do you determine the support reactions in a moments problem?

To determine the support reactions in a moments problem, you must first draw a free-body diagram of the body in question. Then, apply the principle of moments to the body to create an equation with the unknown support reactions. Finally, solve the equation to determine the support reactions.

5. What are the common applications of moments in engineering?

Moments are commonly used in engineering to analyze and design structures such as bridges, buildings, and machines. They are also used in the design of various mechanical systems, such as gears and levers, to ensure that the forces acting on the system are in equilibrium and do not cause any unwanted rotations.

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