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Stella [2.4K]
1 year ago
11

A 1300 kg steel beam is supported by two ropes. (Figure

Physics
1 answer:
Dmitriy789 [7]1 year ago
8 0

Relative to the positive horizontal axis, rope 1 makes an angle of 90 + 20 = 110 degrees, while rope 2 makes an angle of 90 - 30 = 60 degrees.

By Newton's second law,

  • the net horizontal force acting on the beam is

R_1 \cos(110^\circ) + R_2 \cos(60^\circ) = 0

where R_1,R_2 are the magnitudes of the tensions in ropes 1 and 2, respectively;

  • the net vertical force acting on the beam is

R_1 \sin(110^\circ) + R_2 \sin(60^\circ) - mg = 0

where m=1300\,\rm kg and g=9.8\frac{\rm m}{\mathrm s^2}.

Eliminating R_2, we have

\sin(60^\circ) \bigg(R_1 \cos(110^\circ) + R_2 \cos(60^\circ)\bigg) - \cos(60^\circ) \bigg(R_1 \sin(110^\circ) + R_2 \sin(60^\circ)\bigg) = 0\sin(60^\circ) - mg\cos(60^\circ)

R_1 \bigg(\sin(60^\circ) \cos(110^\circ) - \cos(60^\circ) \sin(110^\circ)\bigg) = -\dfrac{mg}2

R_1 \sin(60^\circ - 110^\circ) = -\dfrac{mg}2

-R_1 \sin(50^\circ) = -\dfrac{mg}2

R_1 = \dfrac{mg}{2\sin(50^\circ)} \approx \boxed{8300\,\rm N}

Solve for R_2.

\dfrac{mg\cos(110^\circ)}{2\sin(50^\circ)} + R_2 \cos(60^\circ) = 0

\dfrac{R_2}2 = -mg\cot(110^\circ)

R_2 = -2mg\cot(110^\circ) \approx \boxed{9300\,\rm N}

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Answer:

\boxed{ \bold{ \boxed{ \sf{see \: below}}}}

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Hope I helped!

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6 0
3 years ago
A cannon ball is fired directly upward with a velocity of 160 m/s. How long does it take to fall back to the ground? s How fast
Andrej [43]
To answer this problem, we will use the equations of motions.

Part (a):
For the ball to start falling back to the ground, it has to reach its highest position where its final velocity will be zero.
The equation that we will use here is:
v = u + at where
v is the final velocity = 0 m/sec
u is the initial velocity = 160 m/sec
a is acceleration due to gravity = -9.8 m/sec^2 (the negative sign is because the ball is moving upwards, thus, its moving against gravity)
t is the time that we want to find.
Substitute in the equation to get the time as follows:
v = u + at
0 = 160 - 9.8t
9.8t = 160
t = 160/9.8 = 16.3265 sec
Therefore, the ball would take 16.3265 seconds before it starts falling back to the ground

Part (b):
First, we will get the total distance traveled by the ball as follows:
s = 0.5 (u+v)*t
s = 0.5(160+0)*16.3265
s = 1306.12 meters
The equation that we will use to solve this part is:
v^2 = u^2 + 2as where
v is the final velocity we want to calculate
u is the initial velocity of falling = 0 m/sec (ball starting falling when it reached the highest position, So, the final velocity in part a became the initial velocity here)
a is acceleration due to gravity = 9.8 m/sec^2 (positive as ball is moving downwards)
s is the distance covered = 1306.12 meters
Substitute in the above equation to get the final velocity as follows:
v^2 = u^2 + 2as
v^2 = (0)^2 + 2(9.8)(1306.12)
v^2 = 25599.952 m^2/sec^2
v = 159.99985 m/sec
Therefore, the velocity of the ball would be 159.99985 m/sec when it hits the ground.
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