Can someone help me out, thank you! Figure shows a horizontal uniform beam of ma

Question

Can someone help me out, thank you!

Figure shows a horizontal uniform beam of mass mb and length L that is supported on the left end by a hinge attached to a wall and on the right end by a cable at angle theta with the horizontal. A package of mass mp is positioned on the beam at a distance x from the left end. The total mass is mb + mp = 52.2 kg. Figure gives the tension T in the cable as a function of the package's position as a fraction x/L of the beam length. Find the mass of the package, mp (in kg). (By considering the cases where x = 0 and x = L/2, you can find sin0 and the mass of the beam.)

Explanation / Answer

First place the package at the wall and sum moments about the wall

Here T = 500M

so we have m_b*g*L/2 = T*sin()*L =500*sin()

or m_b*9.8/2 = 500*sin()

which is 4.9*m_b = 500*sin() This is eqn (1)

Now move the package to the center of the beam and sum moments about the wall again

we get m_b*g*L/2 + m_p*g*L/2 = T*sin()*L but here T = 600N

so now we have (m_b +m_p)*g/2 = 600*sin() This is eqn (2)

Now use the fact that m_b + m_p = 60.2 and plug into eqn(2)

so 600*sin() = 52.2*9.8/2 = 255.78

so = arcsin(255.78/600) = 28.03o

Now plug this into eqn(1) to get 4.9*m_b = 500*sin(28.3)

or m_b = 500*sin(29.4)/4.9 = 45.46kg

so m_p = 52.2 - 45.46 = 6.74kg

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