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3 years ago

Plz solve this. plz plz plz plz simple machine

Physics

1 answer:

nalin [4]3 years ago

5 0

Answer:

Explanation:

i. CW moment = 10 N (10 cm) + 30 N (30 cm) - 60 N (40 cm) = - 1400 N-cm

ii. ACW momenet = 60 N (40 cm) - 10 N (10 cm) + 30 N (30 cm) = 1400 N-cm

iii. No. The lever is not balanced in the situation. Because the moment is ± 1400 N-cm. if balance, the moment must be Zero.

iv. the location of 10N by keeping the other loads unchanged to balance the lever is 150 cm

take moment from Δ (support)

60(40) = 10(x) + 30(30)

2400 = 10x + 900

10x = 2400 - 900

10x = 1500

x = 1500/10

x = 150 cm

therefore, the location of 10N by keeping the other loads unchanged to balance the lever is 150 cm

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mel-nik [20]

Answer:

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Given:

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wA₁ = 8π -αA*t₁

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$\theta _{B} =4\pi =\frac{1}{2} \alpha _{B} t_{1}^{2} =\frac{1}{2} (\frac{r_{A} }{r_{B} } )^{3} \alpha _{A} t_{1}^{2}=\frac{1}{2} (\frac{0.15}{0.2} )^{3} \alpha _{A} t_{1}^{2}$

$\alpha _{A} =8\pi (\frac{0.2}{0.15} )^{3} =59.57rad$

$w_{B,1} =\alpha _{B} t_{1}=(\frac{0.15}{0.2} )^{3} \alpha _{A} t_{1}=0.422\alpha _{A} t_{1}$

The velocity at the contact point is equal to:

$v=r_{A} w_{A} =0.15*(8\pi -\alpha _{A} t_{1})=1.2\pi -0.15\alpha _{A} t_{1}$

$v=r_{B} w_{B} =0.2*(0.422\alpha _{A} t_{1})=0.0844\alpha _{A} t_{1}$

Matching both expressions:

$1.2\pi -0.15\alpha _{A} t_{1}=0.0844\alpha _{A} t_{1}\\\alpha _{A} t_{1}=16.09rad/s$

b) The time during which the disks slip is:

$t_{1} =\frac{\alpha _{A} t_{1}^{2}}{\alpha _{A} t_{1}} =\frac{59.574}{16.09} =3.7s$

a) The angular acceleration of each disk is

$\alpha _{A}=\frac{\alpha _{A} t_{1}}{t_{1} } =\frac{16.09}{3.7} =4.35rad/s^{2} (clockwise)$

$\alpha _{B}=(\frac{0.15}{0.2} )^{3} *4.35=1.84rad/s^{2} (clockwise)$

6 0

3 years ago

Plz solve this. plz plz plz plz simple machine ​

Plz solve this. plz plz plz plz simple machine