Block 1 (7 kg) is located on the surface of a table. A hand pushes horizontally

Question

Block 1 (7 kg) is located on the surface of a table. A hand pushes horizontally to the right on block 1 with a normal force of 56 N. The coefficient of kinetic friction between the block and the surface equals 0.4. On a sheet of paper, draw the free body diagram for block 1 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. FORCES on BLOCK 1 Weight force on block 1 by Earth W1E = i + j N Normal force on block 1 by Surface N1S= i + j N Normal force on block 1 by Hand N1H= i + jN Frictional force on block 1 by Surface f1S = i + j N What is the acceleration a of block 1? a = i + j m/s^2

Explanation / Answer

W1E = mg = 7*9.8 = 68.6 N

W1E = 0 i+ 68.6 j N

Force equation in vertical direction is given as

N1s + (56)*sin(0) = W1E

N1s + 0 = 68.6

N1s = 68.6 N

N1S = 0 i + 68.6 j N

frictional force is given as

f1s = 0.4 (N1S ) = 0.4*68.6 = 27.44

f1s = 27.44 i + 0 j

net force is given as

Fnet = T1Rx - f1s = 0 - 27.44 = -27.44 N

acceleration is given as

a = Fnet /m = 27.44 / 7 = 3.2 m/s2

a= 3.2 i + 0 j m/s^2

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