A cable holds a weight of 100 pounds, what is the tension at 30 degrees
1 answer:
The tension in the cable when it is at angle of 30° to the vertical is <u>86.6 lb.</u>
refer to the attached diagram.
The cable has a weight <em>W</em> attached to it and it is at an angle θ 30° to the vertical.
Resolve the weight <em>W</em> into two components W cos θ opposite to the tension <em>T</em> and W sinθ perpendicular to it, as shown in the diagram.
Apply the condition for equilibrium along the direction in which the tension acts.
Substitute 100 lbs for <em>W</em> and 30° for θ.
Thus the tension in the cable when it is at an angle 30° to the vertical is <u>86.6 lb.</u>
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Answer:
a) L=0. b) L = 262 k ^ Kg m²/s and c) L = 1020.7 k^ kg m²/s
Explanation:
It is angular momentum given by
L = r x p
Bold are vectors; where L is the angular momentum, r the position of the particle and p its linear momentum
One of the easiest ways to make this vector product is with the use of determinants
Let's apply this relationship to our case
Let's start by breaking down the speed
v₀ₓ = v₀ cosn 45
voy =v₀ sin 45
v₀ₓ = 9 cos 45
voy = 9 without 45
v₀ₓ = 6.36 m / s
voy = 6.36 m / s
a) at launch point r = 0 whereby L = 0
. b) let's find the position for maximum height, we can use kinematics, at this point the vertical speed is zero
vfy² = voy²- 2 g y
y = voy² / 2g
y = (6.36)²/2 9.8
y = 2.06 m
Let's calculate the angular momentum
L=
L = -px y k ^
L = - (m vox) (2.06) k ^
L = - 20 6.36 2.06 k ^
L = 262 k ^ Kg m² / s
The angular momentum is on the z axis
c) At the point of impact, at this point the height is zero and the position on the x-axis is the range
R = vo² sin 2θ / g
R = 9² sin (2 45) /9.8
R = 8.26 m
L =
L = - x py k ^
L = - x m voy
L = - 8.26 20 6.36 k ^
L = 1020.7 k^ kg m² /s
(a) The velocity of the object on the x-axis is 6 m/s, while on the y-axis is 2 m/s, so the magnitude of its velocity is the resultant of the velocities on the two axes:
And so, the kinetic energy of the object is
(b) The new velocity is 8.00 m/s on the x-axis and 4.00 m/s on the y-axis, so the magnitude of the new velocity is
And so the new kinetic energy is
So, the work done on the object is the variation of kinetic energy of the object:
And so, the kinetic energy of the object is
(b) The new velocity is 8.00 m/s on the x-axis and 4.00 m/s on the y-axis, so the magnitude of the new velocity is
And so the new kinetic energy is
So, the work done on the object is the variation of kinetic energy of the object:
Answer:
a) t = 0.528 s
b) D = 1.62 m
Explanation:
given,
speed of the baseball = 3.75 m/s
angle made with the horizontal = 35°
height of the roof edge = 2.5 m
using equation of motion
4.9 t² + 2.15 t - 2.5 = 0
on solving the above equation
t = 0.528 s
b) D = v cos θ × t
D = 3.75 × cos 35° ×0.528
D = 1.62 m
Answer:
La diferencia entre las tensiones máxima y mínima es de 19.614 newtons.
Explanation:
Puesto que la bola gira en un círculo vertical, existe claramente una diferencia entre las tensiones debido a la influencia de la gravedad y la tensión que resulta de la aceleración centrípeta experimentada por la masa. La máxima tensión ocurre cuando la bola se encuentra en el nadir (o la sima) del trayecto circular, la cual se describe por la Segunda Ley de Newton:
En cambio, la mínima tensión aparece cuando la bola se encuentra en el cénit (o la cima) del trayecto circular, descrita por la misma ley de Newton:
Donde:
,
- Tensiones mínima y máxima, medidas en newtons.
- Masa de la bola, medida en kilogramos.
- Constante gravitacional, medida en metros por segundo al cuadrado.
- Distancia con respecto al eje de rotación, medida en metros.
- Rapidez tangencial, medido en metros por segundo.
Se elimina la aceleración centrípeta de ambas expresiones por igualación:
Ahora, la diferencia entre las tensiones máxima y mínima es:
Si y
, entonces:
La diferencia entre las tensiones máxima y mínima es de 19.614 newtons.