Question

A \(136-\mathrm{kg}\) crate is at rest on the floor. A worker attempts to push it across the floor by applying a 412-N force horizontally. (a) Take the coefficient of static friction between the crate and floor to be \(0.37\) and show that the crate does not move. (b) A second worker helps by pulling up on the crate. What minimum vertical force must this worker apply so that the crate starts to move across the floor? \((c)\) If the second worker applies a horizontal rather than a vertical force, what minimum force, in addition to the original 412-N force, must be exerted to get the crate started?

Short answer

The static frictional force is found to be 494.744 N. (a) Since 494.744 N is more than the 412 N force applied by the worker, the crate does not move. (b) The second worker must apply a minimum vertical force of approximately 1.6 N for the crate to move. (c) If the second worker applies a horizontal force, the minimum additional force needed is approximately 82.74 N.

Step-by-step solution

Calculate the Frictional Force

First calculate the frictional force using the formula: \( F_{friction} = μ . F_N \), where \( F_N \) is the normal force and \( μ \) is the coefficient of friction. Here, \( F_N = mass . g \) where \( g \) is the acceleration due to gravity. Let's take \( g = 9.8 m/s^2 \). Plugging the values, we get \( F_{friction} = 0.37 . 136 . 9.8 = 494.744 N \).

Compare with Applied Force for (a)

Compare the frictional force with the force applied by the worker. The frictional force is 494.744 N and the worker's force is 412 N which is less than the frictional force. Therefore, the crate does not move.

Calculate Vertical Force for (b)

The second worker has to apply enough upward force to lower the frictional force below 412 N. The new normal force will be \( F_{new} = Weight - Vertical\, force \). So, the vertical force = \( F_{new}/μ - Weight \). We want the new frictional force smaller than 412 N, so we set \( F_{new} = 412/0.37 \). Then, we get the vertical force = \( 1120.2 - 136*9.8 = 1.6 N \).

Calculate Additional Horizontal Force for (c)

For the second worker to apply a horizontal force, that force, added to the original 412 N, must be greater than the static friction. So, the additional force needed is \( F_{friction} - 412 = 494.744 - 412 = 82.744 N \).

Key concepts

Coefficient of Static Friction

Understanding the coefficient of static friction is a critical element in the study of physics, particularly when it comes to analyzing the motion of objects. Simply put, it is a dimensionless number that represents how much frictional resistance is present between two surfaces before one begins to slide over the other.

The coefficient of static friction is symbolized by the Greek letter \(\mu_s\), and its value depends on the materials that come into contact. For instance, rubber on concrete would have a higher \(\mu_s\) compared to ice on metal due to the differences in surface textures and properties. In the given problem, the coefficient of static friction between the crate and the floor is \(\mu_s = 0.37\).

This number is crucial because it helps determine the maximum static frictional force that can occur between the crate and the floor. The higher the coefficient, the greater the force required to initiate motion. The interplay of this coefficient with the normal force dictates whether the crate will move or remain stationary when a horizontal force is applied, as seen in the exercise provided.

Normal Force

In physics, the concept of the normal force is pivotal. It represents the force perpendicular to the contact surface that prevents an object from 'falling' into whatever it is resting on. Think of it as the floor pushing upward on your feet as you stand; this is the normal force at play.

The normal force \(F_N\) typically balances the weight of the object in question when it's at rest on a horizontal surface. It is calculated by the mass \(m\) of the object multiplied by the acceleration due to gravity \(g\), which is approximately \(9.8\,m/s^2\) on Earth. For the \(136-kg\) crate, the normal force equals \(136\,kg\) multiplied by \(9.8\,m/s^2\), resulting in a substantial upward force that counters the crate's weight.

However, the scenario gets interesting when an additional vertical force is applied, as in the case of the second worker in the exercise. This additional force effectively alters the normal force, which in turn affects the static friction. The interrelation between these forces becomes apparent when attempting to move the crate.

Applied Force

Applied force is the exertion needed to change an object's state of rest or motion. It can be pushing, pulling, lifting or any other type of force that initiates movement. Illustrated plainly in the exercise, the worker applies a horizontal force of 412 N to set the crate in motion.

This applied force, however, must overcome the static frictional force created by the interaction of the crate with the floor to cause any movement. It is the catalyst for displacement, but its magnitude must be sufficient to surpass the threshold imposed by static friction, which is dependent on both the normal force and the coefficient of static friction.

When an object is in equilibrium, the applied forces are balanced by the reactive forces, such as friction. This is why, in the case of the crate, even though the worker applies a considerable amount of force, if it is not greater than the frictional force, the crate remains immobile.

Frictional Force

Frictional force is that opposing force that surfaces exert to prevent sliding motion. It is always directed opposite to the direction of the applied force trying to move an object. In static friction scenarios, this force increases to match the applied force, up to a particular limit, beyond which slipping occurs.

The amount of frictional force can be computed using the relationship \(F_{friction} = \mu_s . F_N\), which involves the coefficient of static friction and the normal force. In our example, with \(\mu_s = 0.37\) and the normal force calculated from the crate's weight, the frictional force is found to be 494.744 N.

This value is key to understanding why the applied force of 412 N is not enough to budge the crate. It acts as the threshold that the worker's force must exceed to initiate motion. Remember, increasing the normal force or the coefficient of static friction will increase the static frictional force correspondingly, making it harder to slide the object.