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2 years ago

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Two masses hang below a massless meter stick. Mass 1 is located at the 10cm mark with a weight of 15kg, while mass 2 is located

at the 60cm mark with a weight of 30kg. At what point in between the two masses must the string be attached in order to balance the system?

1 answer:

yan [13]2 years ago

7 0

Answer:43.33 cm mark

Explanation:

Given

mass 1 is located at the 10 cm mark with weight of 15 kg

mass 2 is located at 60 cm mark with weight of 30 kg

string should be attached between them to balance the system

so the distance between the the two masses is 50 cm

For system to be balance torque of both the weight must nullify each other

Let us suppose string is at a distance of x cm from 15 kg mass so 30 kg mass is at a distance of 50-x cm

Balancing torque

$15\times x-30\times (50-x)$

$x=\frac{100}{3}=33.33$

so string should be at a mark of 10+33.33=43.33 cm

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Answer:

The tension in the rope is 281.60 N.

Explanation:

Given that,

Length = 3.0 m

Weight = 600 N

Distance = 1.0 m

Angle = 60°

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$Y+F=600$ ....(I)

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On taking moment about base

$X\times l\cos\theta+Y\times l\sin\theta-F\dfrac{l}{2}\sin\theta-T\times d=0$

Put the value into the formula

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$3\cos30 T-T=600\times1.5\sin30-Y \times3\sin30$

$1.598T=450-1.5(600-F)$ ....(III)

We need to calculate the force for ladder

$2F=600\trimes 2$

$F=600\ N$

We need to calculate the tension in the rope

From equation (3)

$1.598T=450-1.5(600-600)$

$1.598T=450$

$T=\dfrac{450}{1.598}$

$T=281.60\ N$

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$891.645=f_r\times 1.5$

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$\mu 100\times 9.8\times \cos 30=594.43$

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