What is units and dimensions? what are dimensions less quantities?
1 answer:
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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
Answer:
3.46 seconds
Explanation:
Since the ball is moving in circular motion thus centripetal force will be acting there along the rope.
The equation for the centripetal force is as follows -
Where, is the mass of the ball,
is the speed and
is the radius of the circular path which will be equal to the length of the rope.
This centripetal force will be equal to the tension in the string and thus we can write,
and,
Thus, m/s.
Now, the total length of circular path = circumference of the circle
Thus, total path length = 2πr = 2 × 3.14 × 2 = 12.56 m
Time taken to complete one revolution = =
= 3.46 seconds.
Thus, the mass will complete one revolution in 3.46 seconds.
Answer:
Explanation:
Coefficients of Friction
Objects in physical contact produce friction which usually manifests as thermal energy being dissipated in the surface where the objects are interacting. It's usually harder to start to move an object from rest, that keeps moving it at a constant speed on the same surface. That is why there are two different coefficients of friction: the static and the dynamic. As mentioned, the static coefficient is greater than the dynamic coefficient
. The car is already moving and is attempting to stop. The coefficient of friction is defined as
Where Fr is the force of friction and N is the normal or the force the road pushes back up on the car. With the given data, we have
The coefficient of friction is dimensionless (doesn't have any units)