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pav-90 [236]
3 years ago
8

? A body is suspended from the ceiling with two wires that make an angle of 40° with the ceiling. The weight of the body is 150N

. What is the tension in each wire?
Physics
1 answer:
andriy [413]3 years ago
4 0
So, the weight is supported by the tension in the two wires, but only part of the tension goes in holding the body: the vertical component.

If the angle between the ceiling and the wire is 40 deg, then the vertical component is

T_y = T * sin(40),

There are two wires, so together they do

2*T*sin(40)

And all this holds the weight of the body, so:

2*T*sin(40)=150 ----> T = 75/sin(40)~116.68 N,

Notice how it is larger than 150N (both wires is ~ 233 N), as the angle becomes smaller, there is less tension that is vertical, so one needs more tension to make the vertical component as large as (half) the weight of the body

Hope it helps!

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Part B) the angular velocity is 195.13 rad/s

Part C)  the angular velocity is 345.913 rad/s

Part D ) the time is t= 7.652 s

Explanation:

Part A) since angular acceleration is related with angular acceleration through:

α = a/R = 10.2 m/s² / 0.23 m =   44.347 rad/s²

Part B) since angular acceleration is related

since

v = v0 + a*(t-t0) =  51.0 m/s + (-10.2 m/s²)*(3.4 s - 2.8 s) = 44.88 m/s

since

ω = v/R = 44.88 m/s/ 0.230 m = 195.13 rad/s

Part C) at t=0

v = v0 + a*(t-t0) =  51.0 m/s + (-10.2 m/s²)*(0 s - 2.8 s) = 79.56 m/s

ω = v/R = 79.56 m/s/ 0.230 m = 345.913 rad/s

Part D ) since the radial acceleration is related with the velocity through

ar = v² / R → v= √(R * ar) = √(0.23 m  * 9.81 m/s²)= 1.5 m/s

therefore

v = v0 + a*(t-t0) → t =(v -  v0) /a + t0 = ( 1.5 m/s - 51.0 m/s) / (-10.2 m/s²) + 2.8 s = 7.652 s

t= 7.652 s

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