Non-Homogenous Seesaw

Classical Mechanics Level 2

A non-homogenous seesaw has a length of 1.5 m. 2 objects with masses m m and M M = 3 kg are put on the seesaw. m m is put 20 cm from the support while M M is put 1.2 m from the support. Calculate the magnitude of the force (in newtons) exerted by the support if the seesaw is in equilibrium. (Round to the nearest integer)

Linear Mass Density for the seesaw : λ ( x ) \lambda(x) = x x k g / m kg/m

Acceleration of Gravity : g = 9.8 g = 9.8 m / s 2 m/s^2


The answer is 261.

Jovan Alfian Djaja by Jovan Alfian Djaja , 18, Indonesia

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1 solution

Let m be x = 0. First, we need to find the centre of mass of the seesaw.

x C O M x_{COM} = ∫ 0 1.5 x 3 d x ∫ 0 1.5 x 2 d x \frac{ \int_0^{1.5} x^3 dx}{ \int_0^{1.5} x^2 dx} = 1 m from m m

The total mass of the seesaw :

m s e e s a w m_{seesaw} = ∫ 0 1.5 x 2 d x \int_0^{1.5} x^2 dx = 9 8 \frac{9}{8} kg

Equilibrium of Torques :

Σ \Sigma τ \tau = 0

m m . 1 5 \frac{1}{5} = 3 . 1.2 + 9 8 \frac{9}{8} . (1-0.2)

m m = 22.5 kg

Equilibrium of Forces :

N = (22.5 + 3 + 9 8 \frac{9}{8} ) kg . 9.8 9.8 m s 2 \frac{m}{s^2}

N ≈ \approx 261 N

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You wrote λ ( x ) = x \lambda(x)=x in the problem. This forced me to see your solution, and I lost the chance to post mine. Be careful while posting a problem.

A Former Brilliant Member - 1 year ago

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