Center of Gravity, objects in eqm

In summary, Two questions were discussed, one involving the calculation of a force applied by a muscle to hold a leg in a specific position, and the other involving the determination of the tension in a support cable and the force exerted on the lower end of a boom. Equilibrium conditions were applied and a trick was suggested for finding the torque. A mistake was made in the calculation for the first question, but was corrected in the end. It was also suggested to solve problems algebraically before plugging in values to avoid errors.
  • #1
Buddha
4
0
Hey guys, I was assigned a bunch of questions by my teacher and I'm having trouble with these two:

A person is sitting with one leg outstretched, so that it makes an angle of 30 degrees with the horizontal, as the drawing indicates. The weight of the leg below the knee is 44.5 N with the center of gravity located below the knee joint. The leg is being held in this position because of force M applied by the quadriceps muscle, which is attached 0.100 m below the knee joint. Obtain the magnitude of M.

A wrecking ball (weight=4800N) is supported by a boom, which may be assumed to be uniform and has a weight of 3600N. As the drawing shows, a support cable runs from the top of the boom to the tractor. The angle between the support cable and the horizontal is 32 degrees, and the angle between the boom and the horizontal is 48 degrees. Find the tension in the support cable and the magnitude of the force exerted on the lower end of the boom by the hinge.

Hopefully the pictures show up...
 

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  • #2
Apply Equilibrium conditions

[tex] \sum_{i}^{n} \vec{F}_{i} = 0 [/tex]

[tex] \sum_{i}^{n} \vec{\tau}_{i} = 0 [/tex]

A good trick for the [tex] \tau [/tex] is selecting an origin which elimnates the most forces, or unknown variables.
 
Last edited:
  • #3
It's been a long time since I've had to do this stuff...
For that first question here is what I've got (let the 44.5 N = X):
ET = My - Xy = 0
My = Xy
Mcos65 = 445cos30
M = 445cos30/cos65
M = 911.89 N
But M should equal 228 N, did I use the wrong angles or am I doing everything wrong?
 
  • #4
You're forgetting

[tex] \vec{\tau} = \vec{F} \times \vec{r} [/tex]

where [itex] |\vec{\tau}| [/itex] is [itex] \tau = |\vec{r}||\vec{F}|sin\theta [/itex]

where theta is the angle between the radius vector and the force vector.

or [itex] \tau = Fd [/itex] where d is perpendicular to the line of action of the force.
 
  • #5
ah crap, thanks, you're right, so it should have been
M = 445cos30(0.25) / [cos65(.1)] = 2279.7 N
But my answer is still off by a multiple of 10, or did I just copy the answer down wrong?
 
  • #6
It's 44.5 N not 445 N.
 
  • #7
haha, oops, I feel like an idiot, thanks for all the help, hopefully I don't make stupid mistakes like this when it's time for a test.
 
  • #8
In my opinion, solve problems algebraicly then plug in the values, it avoids this kind of errors. :smile:
 

1. What is the center of gravity?

The center of gravity is the point at which the entire weight of an object can be considered to act. It is the average location of all the mass in an object.

2. How is the center of gravity determined?

The center of gravity can be determined by finding the point where the object would balance perfectly if suspended. This can be done through mathematical calculations or by physically suspending the object and marking the point where it balances.

3. Does the center of gravity change?

Yes, the center of gravity can change depending on the distribution of mass within an object. For example, if weight is added to one side of an object, the center of gravity will shift towards that side.

4. Why is the center of gravity important?

The center of gravity is important because it helps determine the stability and balance of an object. Objects with a lower center of gravity are typically more stable and less likely to tip over.

5. How does the center of gravity affect objects in equilibrium?

The center of gravity plays a crucial role in determining whether an object is in equilibrium or not. If the center of gravity is directly above the base of support, the object will be in stable equilibrium. If the center of gravity is outside the base of support, the object will not be in equilibrium and will be unstable.

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