Block 1 (m1 = 6 kg) and block 2 (m2 = 7
kg) are adjacent to each other on the surface of a table.
Block 2 is to the LEFT of
block 1. A rope pulls on block 2
up and to the right with a vertical tension of 40 N
upward, and both blocks move right with an
acceleration of magnitude 6
m/s2. The coefficient of kinetic
friction between the blocks and the surface equals
0.4.
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 diagrams, 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 =
36 i + 0 j
N
NET force on Block 2
Fnet2 =
42 i + 0 j
N
FORCES on BLOCK 1
Weight force on block 1 by Earth
W1E = i + j N
Remember that W1E points down (negative y-value) with a magnitude of m1 * g. |
Normal force on block 1 by Surface
N1S = i + j N
Since the block does not move vertically, the upward N1S force must cancel the downward W1E force. |
Frictional force on block 1 by Surface
f1S = i + j N
The kinetic friction on block 1 by the surface (f1S) equals the
kinetic coefficient of friction times the normal force between
block 1 and the surface (N1S). In this particular case with another external force acting on the block (N12), the kinetic friction opposes the direction of motion. |
Normal force on block 1 by block 2
N12 = i + j N
In the x-direction, the sum of the N12 and f1S forces must equal the net force on block 1. |
Incorrect. | Tries 1/2 | Previous Tries |
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
Frictional force on block 2 by Surface
f2S = i + j N
Tension force on block 2 by Rope
T2R = i + j N
Please help me with this
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