What is the correct component force along the axis of the hook?

In summary, the problem asks you to find the component force along the axis of the hook. Once you have found the angle between the resultant force and the axis of the hook, you can calculate the component force using the inverse tangent function.
  • #1
jgreen520
9
0
I worked out the attached problem. I was just looking for a logic check on my work. The problem asks you to find the component force along the axis of the hook.

Thanks
 

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  • #2
Your geometry and trig is off...recheck the angle. If the incline slopes at 4v:3h, then the screw eye slopes at ?
Then properly determine the angle of the screw eye with the horizontal. Your calc for Fy would them be ok, as long as you realize that Fy lies along the axis of the screw eye which is the y axis, not vertically as you have shown.
 
  • #3
Yes you are correct, the angle is supposed to be 53.13 degrees from the horizontal. That makes the angle between the resultant force (150 lbs) and the axis of the eye 28.13 degrees.

However my Fy calculation now seems to give me the wrong answer...I rotated the coordinate system which I think is giving me the problem.
 
  • #4
jgreen520 said:
Yes you are correct, the angle is supposed to be 53.13 degrees from the horizontal. That makes the angle between the resultant force (150 lbs) and the axis of the eye 28.13 degrees.

However my Fy calculation now seems to give me the wrong answer...I rotated the coordinate system which I think is giving me the problem.
you still have the wrong angle. It is given that the slope of the incline is 4v:3h, is that correct? If so, the slope of the screw eye is not the same as the slope of the incline.
 
  • #5
Yes the 4v and 3h is given in the problem. Inverse Tan gives you 53 degrees from the horizontal.
 
  • #6
jgreen520 said:
Yes the 4v and 3h is given in the problem. Inverse Tan gives you 53 degrees from the horizontal.
What I am trying to point out is that the screw eye does not make an angle of 53 degrees with the horizontal, it makes an angle of 53 degrees with the vertical. So the angle it makes with the horizontal is about 37 degrees. Redo the geometry.
 
  • #7
...I found the mistake I was making. Just needed to extend the axis line of the hook to the vertex of the slope triangle to get the correct angle. Which ends up giving me a component force of 147 lbs which is the correct answer.

Thanks!
 

FAQ: What is the correct component force along the axis of the hook?

What is statics?

Statics is a branch of mechanics that deals with the study of forces acting on stationary objects or systems.

2. What is a component force in statics?

A component force is a force that acts in a specific direction, either horizontally or vertically, on an object or a system. It can be broken down into its horizontal and vertical components to better understand its effects.

3. How do you calculate the magnitude of a component force?

The magnitude of a component force can be calculated by using trigonometric functions, such as sine and cosine, to determine the length of the sides of a right triangle formed by the force and its components. The magnitude of the component force is equal to the length of the side opposite to the angle of interest.

4. What is the difference between a component force and a resultant force?

A component force is a single force acting in a specific direction, while a resultant force is the combined effect of multiple forces acting on an object or a system. Resultant forces take into account the magnitude and direction of all the forces acting on an object, while component forces only consider one specific direction.

5. How are component forces used in real-life applications?

Component forces are used in many real-life applications, such as building and bridge design, car mechanics, and aerospace engineering. By understanding the component forces acting on a system, engineers and scientists can ensure the safety and stability of structures and devices.

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