Question

A 3.0 -kg block is at rest on a horizontal floor. If you push horizontally on the 3.0 -kg block with a force of \(12.0 \mathrm{N},\) it just starts to move. (a) What is the coefficient of static friction? (b) A 7.0 -kg block is stacked on top of the 3.0 -kg block. What is the magnitude \(F\) of the force, acting horizontally on the 3.0 -kg block as before, that is required to make the two blocks start to move?

Short answer

(a) The coefficient of static friction is 0.41. (b) The required force is 40.18 N.

Step-by-step solution

Understand the Situation

We have a block that starts to move when a force of 12 N is applied, which means this is the force needed to overcome the static friction. This force relates to the equation for static friction: \[ f_s = \mu_s N \] where: - \(f_s\) is the static friction force, - \(\mu_s\) is the coefficient of static friction, and - \(N\) is the normal force.

Calculate Normal Force for One Block

Since the block is at rest and on a horizontal surface, its normal force \(N\) is equal to its weight. The weight is calculated as:\[ N = m \, \cdot g \]For the 3.0 kg block, this is:\[ N = 3.0 \, \text{kg} \cdot 9.8 \, \text{m/s}^2 = 29.4 \, \text{N} \]

Find the Coefficient of Static Friction (\(\mu_s\))

Using the static friction formula: \[ f_s = \mu_s N \]and using the values \(f_s = 12 \, \text{N}\) and \(N = 29.4 \, \text{N}\), we can find \(\mu_s\): \[ 12 \, \text{N} = \mu_s \cdot 29.4 \, \text{N} \] \[ \mu_s = \frac{12}{29.4} \approx 0.41 \]

Combine Masses and Calculate New Normal Force

When a 7.0 kg block is added on top of the 3.0 kg block, the total mass becomes \(10.0 \, \text{kg}\). Calculate the new normal force:\[ N = (3.0 + 7.0) \, \text{kg} \, \cdot 9.8 \, \text{m/s}^2 = 98.0 \, \text{N} \]

Calculate New Static Friction Force

With the combined blocks and same static friction coefficient, calculate the force needed to move them:\[ f_s = \mu_s N \]\[ f_s = 0.41 \, \cdot 98.0 \, \text{N} \approx 40.18 \, \text{N} \]

Conclusion

The coefficient of static friction for the first part of the problem is 0.41. The force required to move both blocks is approximately 40.18 N. The calculations show that static friction increases with the weight of the objects in contact.

Key concepts

Static Friction

Static friction acts between two surfaces that aren't moving relative to each other. It's the force you need to overcome to start moving an object at rest. Think about when you try to push a heavy box that's sitting on the floor. At first, it doesn't move. That's because of static friction holding it in place.
The static friction force (fs) can be calculated by the equation:\[ f_s = \mu_s \cdot N \]

• \( f_s \) is the static friction force.
• \( \mu_s \) is the coefficient of static friction.
• \( N \) is the normal force.

In the given problem, to find when the object just starts moving, the static friction force equals the applied force, here 12 N. Knowing the force required to start the motion helps us calculate the coefficient of static friction (\mu_s). When a heavier object is placed on the block, as in the example with added mass, the static friction increases because the normal force increases.

Normal Force

The normal force is the support force exerted by a surface to keep an object resting on it from falling. It's always perpendicular to the surface. Let's paint a mental picture: when an apple rests on a table, the table pushes back up on the apple with the same force that gravity is pulling down on it. This upward force is the normal force.

In our exercise, the normal force is calculated as the weight of the block, which is the mass of the block multiplied by the gravitational acceleration. For the 3-kg block:\[ N = m \cdot g = 3.0 \, \text{kg} \times 9.8 \, \text{m/s}^2 = 29.4 \, \text{N} \]

This normal force is crucial because it directly affects the amount of static friction. Adding a second block increases the total weight, leading to a larger normal force, which in turn increases the static friction.

Newton's Laws of Motion

Newton's laws describe how objects move and interact. The first law is particularly relevant here, often called the "law of inertia." It states that an object will remain at rest or move at a constant velocity unless acted upon by a net force.

In this exercise, the block remains stationary until the applied horizontal force exceeds the static friction. When you apply exactly 12 N, the block just starts moving, meaning the forces are balanced until that point. Increasing the mass with an additional block requires recalibration of the forces involved according to Newton's third law — action and reaction.

Newton’s second law, which is summed up as \( F = ma \), although not directly calculated in this problem, is inherently a guiding principle in understanding how the forces interact once the block begins moving. When the frictional force is conquered, it's due to an unbalanced force leading to acceleration of the objects.