The weightless horizontal bar in the figure below is in equilibrium. Scale B reads 4.00 kg (N.B. as you know, the scale should read N, but no one told the manufacturer). The distances in the figure (which is not to scale) are: D1 = 9.0 cm, D2 = 8.0 cm, and D3 = 6.5 cm. The mass of block X is 0.96 kg and the mass of block Y is 1.93 kg.
2 answers
without the figure, I have no idea what the distances mean.
I believe that I know this figure.
If it is so, then...
Let's examine the torques. The torque on A must equal the torque on B
since the system is at rest. The torque on B can be found by summing the
torques from mass x and mass y. Since mass z is on the axis of rotation (at B),it produces no torque.
M(B) = (D2 + D3) •m(x) + D3•m(y),
The torque on A can be calculated by referencing it from B. In other words,
what torque must be applied at A to counteract the torques from m(x) and m(y)? This can by found by denoting the variable m(A) to represent the mass (the scale reading) at A to counteract the torque calculated above. That is,
m(A)•(D1+D2+D3) = (D2 + D3) •m(x) + D3•m(y),
Thus, the scale reading at A, that is, m(A)
m(A) = [(D2 + D3) •m(x) + D3•m(y)]/ (D1+D2+D3)
If it is so, then...
Let's examine the torques. The torque on A must equal the torque on B
since the system is at rest. The torque on B can be found by summing the
torques from mass x and mass y. Since mass z is on the axis of rotation (at B),it produces no torque.
M(B) = (D2 + D3) •m(x) + D3•m(y),
The torque on A can be calculated by referencing it from B. In other words,
what torque must be applied at A to counteract the torques from m(x) and m(y)? This can by found by denoting the variable m(A) to represent the mass (the scale reading) at A to counteract the torque calculated above. That is,
m(A)•(D1+D2+D3) = (D2 + D3) •m(x) + D3•m(y),
Thus, the scale reading at A, that is, m(A)
m(A) = [(D2 + D3) •m(x) + D3•m(y)]/ (D1+D2+D3)
1 answer