A block of mass M is pulled on a smooth horizontal table
by a string making an angle 9 with the horizontal as
shown in figure (5.7). If the acceleration of the block is
a, find the force applied by the string and by the table
on the block
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We have a block of mass M on a smooth horizontal table. The block is pulled by a string that makes an angle theta with the horizontal. The block accelerates with acceleration a. We need to find the forces applied by the string and by the table on the block.
Now let's analyze the forces acting on the block. We have three forces: the tension T from the string, the weight Mg acting downward, and the normal force N from the table acting upward. The key step is to resolve the tension force into its components. The horizontal component is T cosine theta, and the vertical component is T sine theta.
Now we apply Newton's second law in the horizontal direction. Since the block accelerates horizontally with acceleration a, the net horizontal force equals mass times acceleration. The only horizontal force is T cosine theta, so we have T cosine theta equals M a. Solving for the tension T, we get T equals M a divided by cosine theta.
In the vertical direction, the block has no acceleration, so the net vertical force is zero. We have the normal force N upward, the vertical component T sine theta upward, and the weight Mg downward. Setting the sum equal to zero: N plus T sine theta minus Mg equals zero. Substituting our expression for T and simplifying, we get N equals Mg minus Ma tangent theta.
In summary, we have found the two forces acting on the block. The force applied by the string is T equals M a divided by cosine theta. The force applied by the table is the normal force N equals M g minus M a tangent theta. Notice that as the angle theta increases, the string force increases while the normal force from the table decreases. This completes our force analysis of the block on the inclined string.