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a_lawson_2k

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Hi, got stumped again, but the sooner I figure out how to do these, the easier the others will be in the future.

for the 240N weight

[tex]\SigmaF_x=0=T-240N[/tex]

so we know what the tension is for the rope. for the large pulley, going clockwise:

[tex]\Sigma\Tau=0=(0.25m)(240N)-(1)(W)[/tex] so the weight was pretty simple to calculate (60N)

The second part is where complications arose; the second equation became:

[tex]\Sigma\Tau=0=(0.25m)(240N)-(1)(60N)(cos(\theta))-(2)(20n)(cos(\theta))[/tex]

I solved for theta and got 55.7 degrees, the book said 53.2...

For a separate problem, I need clarification about what the definitions for slipping and tipping are in the context of equilibrium, and I'm pretty sure I can take it from there.

I got as far as [tex]\SigmaF_x=0=F_1+F_2-1960N[/tex]

two variables, the equation for torque (counterclockwise) about the center of mass should handle this:

[tex]\Sigma\Tau=0=(P_2)x(F_2)-(P_1)x(F_1)=F_2\frac{\sqrt{2}}{2}-F_1\frac{\sqrt{2}}{4}[/tex]

This doesn't get the results I need (1370N for F1 and 590N), so I think I'm computing the torque wrong, since the answers both add to 1960 (thank goodness!)

I labeled the tension, natural force exerted from the pivot, and weight of the beam.

[tex]\Sigma\F_x=n-Tcos(\theta), \Sigma\F_y=Tsin(\theta)-w[/tex]

Since the girder can't accelerate in the X direction, the point cannot be closer to the pivot on the girder than L/2.

From there, the torque about the pivot would be

[tex]\Sigma\Tau=(\lambda+\frac{L}{2})T* arctan(\frac{\lambda}{h})-\frac{L}{2}W[/tex]

The answer was L/2+h^2/L, and I couldn't get to that answer. Not sure how to go from here...

Thanks again for the help

11-59 said:A circular disc 0.5m in diameter pivoted about a horizontal axis through its center, has a cord wrapped around its rim. The cord passed over a frictionless pulley P and is attached to an object weighing 240N. Uniform rod 2m long is fastened to the disc, with one end at the center of the disc. The apparatus is in equilibrium, with the rod horizontal

a) what is the weight of the rod?

b) what is the new equilibrium direction of the rod when a second object weighing 20N is suspended from the other end, as http://img159.imageshack.us/img159/5170/1159ij7.gif" ? (what angle does it make with the horizontal?)

for the 240N weight

[tex]\SigmaF_x=0=T-240N[/tex]

so we know what the tension is for the rope. for the large pulley, going clockwise:

[tex]\Sigma\Tau=0=(0.25m)(240N)-(1)(W)[/tex] so the weight was pretty simple to calculate (60N)

The second part is where complications arose; the second equation became:

[tex]\Sigma\Tau=0=(0.25m)(240N)-(1)(60N)(cos(\theta))-(2)(20n)(cos(\theta))[/tex]

I solved for theta and got 55.7 degrees, the book said 53.2...

For a separate problem, I need clarification about what the definitions for slipping and tipping are in the context of equilibrium, and I'm pretty sure I can take it from there.

11-61 said:Two people are carrying a 200kg crate up a flight of stairs; the crate is 1,25m long and 0.5m high, and its center of gravity is at its geometric center. The stairs make a 45 degree angle with the floor, and the crate is carried at a 45 degree angle so that it is parallel with the stairs. If the force from each person is vertical, what is the magnitude of each of the forces? http://img109.imageshack.us/img109/5865/1961cp8.gif"

I got as far as [tex]\SigmaF_x=0=F_1+F_2-1960N[/tex]

two variables, the equation for torque (counterclockwise) about the center of mass should handle this:

[tex]\Sigma\Tau=0=(P_2)x(F_2)-(P_1)x(F_1)=F_2\frac{\sqrt{2}}{2}-F_1\frac{\sqrt{2}}{4}[/tex]

This doesn't get the results I need (1370N for F1 and 590N), so I think I'm computing the torque wrong, since the answers both add to 1960 (thank goodness!)

11-83 said:Heavy horizontal girder of length L has several objects suspended from it, it is supported by a tensionless pivot at its left end and a cable of negligible weight attached to an I-beam at a point distance h directly above the girder's center. Where should the other end be placed to minimize tension of the cable? HINT: remember to the maximum distance from pivot to the end of the beam is L. http://img171.imageshack.us/img171/9053/1983kl9.gif"

I labeled the tension, natural force exerted from the pivot, and weight of the beam.

[tex]\Sigma\F_x=n-Tcos(\theta), \Sigma\F_y=Tsin(\theta)-w[/tex]

Since the girder can't accelerate in the X direction, the point cannot be closer to the pivot on the girder than L/2.

From there, the torque about the pivot would be

[tex]\Sigma\Tau=(\lambda+\frac{L}{2})T* arctan(\frac{\lambda}{h})-\frac{L}{2}W[/tex]

The answer was L/2+h^2/L, and I couldn't get to that answer. Not sure how to go from here...

Thanks again for the help

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