A block of mass M is pulled on a smooth horizontal table by a string making an angle 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. Show with diagram

1. Draw a Free-Body Diagram (FBD)

• Block of mass $M$ sits on a smooth horizontal table (so friction $=0$).
• Forces acting on the block:
  1. Weight: $Mg$ downward.
  2. Normal reaction by the table: $N$ upward.
  3. Tension in the string: $T$ acting at an angle $\theta$ above the horizontal.

2. Resolve Tension into Components

Because $T$ is slanted:

• Horizontal component: $T\cos\theta$ (causes acceleration $a$)
• Vertical component: $T\sin\theta$ (helps lift the block)

3. Apply Newton’s Second Law Along Each Axis

Horizontal (x-axis):

$\text{Net force}_x = M a$

So,

$T\cos\theta = Ma \quad\Rightarrow\quad T = \frac{Ma}{\cos\theta}$

Vertical (y-axis): The block doesn’t leave the table, so vertical acceleration is zero.

$\text{Upward forces} = \text{Downward forces}$

$N + T\sin\theta = Mg$

Substitute $T$:

$N = Mg - T\sin\theta = Mg - \left(\frac{Ma}{\cos\theta}\right)\sin\theta = Mg - Ma\tan\theta$

Thus:

• Tension $T = \dfrac{Ma}{\cos\theta}$
• Normal reaction $N = Mg - Ma\tan\theta$

4. Physical Meaning

• Higher acceleration ($a$) or larger angle ($\theta$) demands a bigger pull.
• Pulling more upward ($\theta$ large) reduces the normal force since you partly support the block’s weight.

Imagine pulling a toy car on a very smooth floor

1. The floor is so smooth that it offers no friction at all.
2. You tie a thread to the car and pull it a little upward, not straight along the floor.
3. Because you pull slanting upward, part of your pull drags the car forward and part of it lifts the car slightly.
4. The car begins to move with some acceleration $a$.
5. We want to know two things:
   • How strong is your pull (the string force)?
   • How hard is the floor pushing up on the car (the normal force)?

To find this, we simply split your pull into two parts: a forward part and an upward part, then use Newton’s second law ($F = ma$).

Step-by-Step Calculation

1. Free-Body Diagram

   • Weight $Mg$ downward.
   • Normal force $N$ upward from table.
   • Tension $T$ at angle $\theta$ above horizontal.

2. Resolve Tension

   $T_x = T\cos\theta$

   $T_y = T\sin\theta$

3. Apply Newton’s Laws
   Horizontal:

   $T\cos\theta = Ma \quad (1)$

   Vertical (no vertical motion):

   $N + T\sin\theta = Mg \quad (2)$

4. Find Tension $T$
   From (1):

   $T = \frac{Ma}{\cos\theta}$

5. Find Normal Force $N$
   Plug $T$ into (2):

   $N = Mg - T\sin\theta$

   $N = Mg - \left(\frac{Ma}{\cos\theta}\right)\sin\theta$

   $N = Mg - Ma\tan\theta$

Final Answers

• Tension in string:
  $T = \dfrac{Ma}{\cos\theta}$

• Normal force by table:
  $N = Mg - Ma\tan\theta$