Question 1

15. In the following figure, assume the mass of the meter stick is negligible and thatthe system is in equilibrium about the pivot point “S”. Find the mass of m3 if themasses of mı and m2 are 3.0 kg and 2.5 kg respectfully.30 cm +30 cm40 cmto 30 cmTSm,b.09m22.349m32kg -

15. In the following figure, assume the mass of the meter stick is negligible and-example-1

Question 2

We will have the following:

$(3.0\operatorname{kg}\cdot9.8m/s^2)(0.7m)+(2.5\operatorname{kg}\cdot9.8m/s^2)(0.4m)=F_(m3)\cdot0.3m$
$\Rightarrow20.58N\cdot m+9.8N\cdot m=F_(m3)\cdot0.3m\Rightarrow30.38N\cdot m=F_(m3\cdot0.3m)$
$\Rightarrow F_(m3)=(30.38N\cdot m)/(0.3m)\Rightarrow F_(m3)=(1519)/(15)N$
$\Rightarrow m_3=(((1519)/(15)N))/(9.8m/s^2)\Rightarrow m_3=(31)/(3)kg$
$\Rightarrow m_3=10.3333333\ldots kg\Rightarrow m_3\approx10.33\operatorname{kg}$

So, m3 will be exactly 31/3 kg, that is approximately 10.33 kg.

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