miktalmyers
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Calculating mass of a block...
Force equations/Kinetic equations
FA = mA * 9.8 m/sec2
FA = FB
where FB is the force acting through the string from A to B
The forces on B are friction from the table, friction from C, and the force it takes to move B's mass at 0.8 m/sec2
FB = FfB + FfC + (1.4 kg * 0.8 m/sec2)
FfC = 0.3 * ( 0.6 kg * 9.8 m/sec2 )
FfC = 1.764 N
FfB = 0.25 * [ (1.4 kg + 0.6 kg) * 9.8m/sec2 ]
FfB = 4.9 N
FB = 1.764 N + 4.9 N + (1.4 kg * 0.8 m/sec2)
FB = 7.784 N
FA = 7.784 N
7.784 N = mA * 9.8 m/sec2
mA = 0.794286 kg ~ 794 grams
Homework Statement
Homework Equations
Force equations/Kinetic equations
The Attempt at a Solution
FA = mA * 9.8 m/sec2
FA = FB
where FB is the force acting through the string from A to B
The forces on B are friction from the table, friction from C, and the force it takes to move B's mass at 0.8 m/sec2
FB = FfB + FfC + (1.4 kg * 0.8 m/sec2)
FfC = 0.3 * ( 0.6 kg * 9.8 m/sec2 )
FfC = 1.764 N
FfB = 0.25 * [ (1.4 kg + 0.6 kg) * 9.8m/sec2 ]
FfB = 4.9 N
FB = 1.764 N + 4.9 N + (1.4 kg * 0.8 m/sec2)
FB = 7.784 N
FA = 7.784 N
7.784 N = mA * 9.8 m/sec2
mA = 0.794286 kg ~ 794 grams