Statics - finding a moment of a force in xyz plane

• yugeci
In summary, the conversation discusses finding the moment of a force using the equations M = r x F and M = F x d, and resolving forces using Fx/Fy/Fz = F cos/sin theta. The conversation also addresses an error in calculating the vector R and provides an alternative method for finding the torque using a unit vector perpendicular to the plane formed by the force and distance. The conversation concludes with a clarification on finding the perpendicular distance in the equation for torque.
yugeci

Homework Statement

I attached the question.

Homework Equations

M = r x F
M = F x d
Resolving forces, Fx/Fy/Fz = F cos/sin theta (depending on which angle you take)

The Attempt at a Solution

I can easily find the moment of the 100 N and 120 forces, but I'm having trouble finding the moment of the 80 N force. I know we use r x F, but getting the r and F vectors is a problem for me. Here is my attempt:

Fx = -80 cos 20 = -75.18
Fy = 80 sin 20 cos 60 = 13.68
Fz = 80 sin 20 sin 60 = 23.70

F = (-75.18i + 13.68j + 23.70k)

Rx = 0, to get Ry and Rz I made a triangle with the 180mm length and 30 degree angle.

Ry = 180 cos 30 = 155.88
Rz = -180 sin 30 = -90

R = (0, 155.88, -90)

M = r x F

When I take the cross product I get = { 4923.9972, 6766.2, 11719.0584 }

While the actual answer is { 0, -6770, 1172 }

Where did I make the mistake?

Attachments

• toughone.png
10.3 KB · Views: 1,177
Your vector R seems incorrect. It is easier if you remember that the torque is perpendicular to the plane formed by F and R. find the unit vector perpendicular to the plane and multiply it by the magnitude of the torque.

dauto said:
Your vector R seems incorrect. It is easier if you remember that the torque is perpendicular to the plane formed by F and R.
I had a feeling my R vector was wrong.. how would you calculate it correctly with the method I did then? I thought resolving the 180mm into y and z components would do it.

find the unit vector perpendicular to the plane and multiply it by the magnitude of the torque.
This seems to be what the solution says.

M2 = 80(0.180 cos 20) (-j sin 30 - k cos 30)

My question is, for the torque aren't the force and distance supposed to be perpendicular? If so how can you write 80(0.180 cos 20)? The force would be along the x-axis, but the 180mm length doesn't look perpendicular to it.

You method should work too. I think you just got some of the components with the wrong sign.

yugeci said:
My question is, for the torque aren't the force and distance supposed to be perpendicular? If so how can you write 80(0.180 cos 20)? The force would be along the x-axis, but the 180mm length doesn't look perpendicular to it.
That's what the cos(20) is for. The perpendicular distance is 0.180 cos (20); or, equivalently, the component of the 80 force perpendicular to the 0.18 measurement is 80 cos 20.

1. What is a moment of a force?

The moment of a force is a measure of the tendency of the force to rotate an object about a specific point or axis. It is also known as torque.

2. How is the moment of a force calculated?

The moment of a force is calculated by multiplying the magnitude of the force by the perpendicular distance from the point of rotation to the line of action of the force.

3. What is the difference between a positive and negative moment?

A positive moment indicates that the force is creating a clockwise rotation, while a negative moment indicates a counterclockwise rotation. The direction of the rotation is dependent on the choice of axis or point of rotation.

4. Can the moment of a force be zero?

Yes, the moment of a force can be zero if the force is applied at the point of rotation or if the force is parallel to the line of action passing through the point of rotation.

5. How is the moment of a force affected by the distance from the point of rotation?

The moment of a force increases as the distance from the point of rotation increases. This means that the farther the force is from the point of rotation, the greater its tendency to cause rotation.

• Introductory Physics Homework Help
Replies
41
Views
681
• Introductory Physics Homework Help
Replies
8
Views
518
• Introductory Physics Homework Help
Replies
2
Views
806
• Introductory Physics Homework Help
Replies
3
Views
4K
• Introductory Physics Homework Help
Replies
8
Views
374
• Introductory Physics Homework Help
Replies
5
Views
972
• Introductory Physics Homework Help
Replies
10
Views
2K
• Introductory Physics Homework Help
Replies
4
Views
2K
• Introductory Physics Homework Help
Replies
22
Views
2K
• Introductory Physics Homework Help
Replies
5
Views
6K