Animated Solution for Physics - Laws of Motion: A block of mass $5\text{ kg}$ is (i) pushed in case (A) and (ii) pulled in case (B), by a force $F=20\text{ N}$, making an angle of $30^{\circ }$ with the horizontal, as shown in the figures. The coefficient of friction between the block, the floor is $\mu =0.2$. The difference between the accelerations of the block, in case (B) and case (A) will be (Take, $g=10{\text{ ms}}^{-2}$)

The physics of pushing versus pulling is something you experience every day. Have you ever noticed that dragging a heavy suitcase behind you is significantly easier than trying to push it forward? This classic JEE problem mathematically proves exactly why that happens. It all comes down to how the angle of your applied force manipulates the normal force, and consequently, the friction.

Case A (Pushing)

Let's start by looking at Case A, where the block is being pushed by a force $F=20\text{ N}$ at a downward angle of $30^{\circ }$.

When you push down at an angle, your force has two components. The horizontal component, $F\cos 30^{\circ }$, is what actually tries to move the block forward. However, the vertical component, $F\sin 30^{\circ }$, pushes the block into the floor.

Because the block is being pressed harder into the floor, the floor must push back with a greater normal force ($N_A$) to support it. The normal force must balance both the weight of the block ($mg$) and this downward push:

Substituting our values ($m=5\text{ kg}$, $g=10{\text{ ms}}^{-2}$, $F=20\text{ N}$):

Since kinetic friction is directly proportional to the normal force ($f=\mu N$), a higher normal force means higher friction. With $\mu =0.2$:

Now, we can find the acceleration $a_1$ using Newton's Second Law. The net horizontal force is the driving force minus the friction:

Case B (Pulling)

Now, let's shift our focus to Case B, where the block is being pulled. The force $F$ is still $20\text{ N}$ at $30^{\circ }$, but now it points upwards and away from the block.

This upward angle is a game changer. The vertical component, $F\sin 30^{\circ }$, is now acting upwards, effectively lifting the block slightly. Because the block is being lifted, it doesn't press as hard against the floor. The normal force ($N_B$) required from the ground decreases:

With a smaller normal force, the friction force also drops significantly:

This is the mathematical proof of why pulling is easier! The opposing friction is only $8\text{ N}$ compared to the $12\text{ N}$ when pushing. Let's calculate the new acceleration $a_2$:

The problem asks for the difference between the two accelerations, $\Delta a=a_2-a_1$.

Notice how we kept the acceleration values in their exact fractional forms. This was a strategic move to avoid messy decimal approximations. When we subtract them, the irrational $\sqrt{3}$ terms beautifully cancel out:

Whenever you face a mechanics problem involving angled forces, always pay close attention to the vertical components. They dictate the normal force, which dictates the friction, which ultimately decides how the object moves!