Block 1 (m1 = 6 kg) is resting on top of block 2 (m2 = 2 kg) on the floor of an elevator. The elevator is moving downward with an increasing speed and the magnitude of the acceleration is 1 m/s2. On a sheet of paper, draw the free body diagrams for block 1 and block 2 using the two-subscript notation from class. After completing the free body diagram, enter below each force and its x & y-components. Remember that the x-component is the "i" component and the y-component is the "j" component. NET force on Block 1 Fnet1 = i + j N NET force on Block 2 Fnet2 = i + j N Tries 0/2 FORCES on BLOCK 1 Weight force on block 1 by Earth W1E = i + j N Normal force on block 1 by block 2 N12 = i + j N Tries 0/2 FORCES on BLOCK 2 Weight force on block 2 by Earth W2E = i + j N Normal force on block 2 by block 1 N21 = i + j N Normal force on block 2 by Surface N2S = i + j N
Block 1 (m1 = 6 kg) is resting on top of block 2 (m2 = 2 kg) on the floor of an elevator. The elevator is moving downward with an increasing speed and the magnitude of the acceleration is 1 m/s2. On a sheet of paper, draw the free body diagrams for block 1 and block 2 using the two-subscript notation from class. After completing the free body diagram, enter below each force and its x & y-components. Remember that the x-component is the "i" component and the y-component is the "j" component.
NET force on Block 1
Fnet1 = i + j N
NET force on Block 2
Fnet2 = i + j N
| Tries 0/2 |
FORCES on BLOCK 1
Weight force on block 1 by Earth
W1E = i + j N
Normal force on block 1 by block 2
N12 = i + j N
| Tries 0/2 |
FORCES on BLOCK 2
Weight force on block 2 by Earth
W2E = i + j N
Normal force on block 2 by block 1
N21 = i + j N
Normal force on block 2 by Surface
N2S = i + j N
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