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2 years ago

The force F shown in Figure 4.30 has a moment of 40 Nm about the pivot. Calculate the magnitude

Physics

1 answer:

salantis [7]2 years ago

8 0

$\boxed{\sf \tau=rFsin\theta}$

Put values

$\\ \rm\hookrightarrow 40=2Fsin40$

$\\ \rm\hookrightarrow Fsin40=20$

$\\ \rm\hookrightarrow 0.64F=20$

$\\ \rm\hookrightarrow F=31.25N$

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melomori [17]

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Given the data in the question;

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300 = 0 + 22500 × t

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F = ma

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