Answer:
126.99115 g
Explanation:
50 g at 90 cm
Stick balances at 61.3 cm
x = Distance of the third 0.6 kg mass
Meter stick hanging at 50 cm
Torque about the support point is given by (torque is conserved)
The mass of the meter stick is 126.99115 g
Which statement best compares the momentum of a 5-kilogram fish swimming at a speed of 10 meters/second and a 2-kilogram fish sw
2 answers:
Answer:
The momentum of the 5-kilogram fish is greater than the momentum of the 2-kilogram fish
Explanation:
The momentum of each fish can be calculated as follows:
where
m is the mass of the fish
v is the velocity of the fish
- For the first fish, we have:
- For the second fish:
Therefore, the momentum of the first fish is greater than the second fish.
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Explanation:
Draw a free body diagram of the bar.
There are 3 reaction forces at O in the x, y, and z direction (Ox, Oy, and Oz).
There is a tension force Tac at A in the direction of the rope. There are also tension forces Tbd and Tbe at B in the direction of the ropes.
Finally, there is a weight force mg pulling down halfway between A and B, where m = 400 kg.
There are 6 unknown variables, so we'll need 6 equations to solve. Summing the forces in the x, y, and z direction will give us 3 equations. Summing the torques about the x, y, and z axes will give us 3 more equations.
First, let's find the components of the tension forces.
Tbe is purely in the z direction.
Tbd has components in the y and z directions. The length of Tbd is √8.
(Tbd)y = 2/√8 Tbd
(Tbd)z = 2/√8 Tbd
Tac has components in the x, y, and z directions. The length of Tac is √6.
(Tac)x = 1/√6 Tac
(Tac)y = 1/√6 Tac
(Tac)z = 2/√6 Tac
Sum of the forces in the +x direction:
∑F = ma
Ox − (Tac)x = 0
Ox − 1/√6 Tac = 0
Sum of the forces in the +y direction:
∑F = ma
Oy + (Tac)y + (Tbd)y − mg = 0
Oy + 1/√6 Tac + 2/√8 Tbd − mg = 0
Sum of the forces in the +z direction:
∑F = ma
Oz − (Tac)z − (Tbd)z − Tbe = 0
Oz − 2/√6 Tac − 2/√8 Tbd − Tbe = 0
Sum of the torques counterclockwise about the x-axis:
∑τ = Iα
mg (2 m) − (Tac)y (2 m) − (Tbd)y (2 m) = 0
mg − (Tac)y − (Tbd)y = 0
mg − 1/√6 Tac − 2/√8 Tbd = 0
Sum of the torques counterclockwise about the y-axis:
∑τ = Iα
-(Tac)x (2 m) + (Tbd)z (1.5 m) + Tbe (1.5 m) = 0
-4 (Tac)x + 3 (Tbd)z + 3 Tbe = 0
-4/√6 Tac + 6/√8 Tbd + 3 Tbe = 0
Sum of the torques counterclockwise about the z-axis:
∑τ = Iα
-mg (0.75 m) + (Tbd)y (1.5 m) = 0
-mg + 2 (Tbd)y = 0
-mg + 4/√8 Tbd = 0
As you can see, by summing the torques about axes passing through O, we were able to write 3 equations independent of those reaction forces. We can solve these equations for the tension forces, then go back and find the reaction forces.
-mg + 4/√8 Tbd = 0
4/√8 Tbd = mg
Tbd = √8 mg / 4
Tbd = √8 (400 kg) (9.8 m/s²) / 4
Tbd = 2772 N
mg − 1/√6 Tac − 2/√8 Tbd = 0
1/√6 Tac = mg − 2/√8 Tbd
Tac = √6 (mg − 2/√8 Tbd)
Tac = √6 ((400 kg) (9.8 m/s²) − 2/√8 (2772 N))
Tac = 4801 N
-4/√6 Tac + 6/√8 Tbd + 3 Tbe = 0
3 Tbe = 4/√6 Tac − 6/√8 Tbd
Tbe = (4/√6 Tac − 6/√8 Tbd) / 3
Tbe = (4/√6 (4801 N) − 6/√8 (2772 N)) / 3
Tbe = 653 N
Now, using our sum of forces equations to find the reactions:
Ox − 1/√6 Tac = 0
Ox = 1/√6 Tac
Ox = 1/√6 (4801 N)
Ox = 1960 N
Oy + 1/√6 Tac + 2/√8 Tbd − mg = 0
Oy = mg − 1/√6 Tac − 2/√8 Tbd
Oy = (400 kg) (9.8 m/s²) − 1/√6 (4801 N) − 2/√8 (2772 N)
Oy = 0 N
Oz − 2/√6 Tac − 2/√8 Tbd − Tbe = 0
Oz = 2/√6 Tac + 2/√8 Tbd + Tbe
Oz = 2/√6 (4801 N) + 2/√8 (2772 N) + 653 N
Oz = 6533 N
