Question

A 12 -kg block of steel is at rest on a horizontal table. The coefficient of static friction between block and table is \(0.52 .(a)\) What is the magnitude of the horizontal force that will just start the block moving? ( \(b\) ) What is the magnitude of a force acting upward \(62^{\circ}\) from the horizontal that will just start the block moving? \((c)\) If the force acts down at \(62^{\circ}\) from the horizontal, how large can it be without causing the block to move?

Short answer

The magnitude of the horizontal force that will just start the block moving is approximately \( 61.152N \). For the force acting upwards \( 62^\circ \) from the horizontal, you have to perform a simultaneous equation to find the magnitude. The magnitude of the force acting downwards \( 62^\circ \) from the horizontal can be obtained by solving the equation from Step 3.

Step-by-step solution

Evaluate the horizontal force

To find the magnitude of the horizontal force that will just start the block moving, use the equation for static friction. First, calculate the normal force \( F_n \) which equals the product of mass (\( m \)) and gravity (\( g \)), i.e. \( F_n = m*g = 12kg * 9.8m/s^2 = 117.6N \). The maximum static friction that will start the block moving can be calculated by \( f_s = \mu_s * F_n = 0.52 * 117.6N = 61.152N \). So this force is the answer to part \( a \).

Calculate upward force

For part \( b \), you need to calculate the magnitude of an upward force from the horizontal that will just cause the block to move. This force must balance the force of gravity and the static friction. Let's denote this force as \( F \). Since the force \( F \) makes an angle \( 62^\circ \) with the horizontal, the horizontal and vertical components are \( F \cos 62^\circ \) and \( F \sin 62^\circ \) respectively. Thus, we have the following equations: \( F \cos 62^\circ = f_s = 61.152N \) (balances the static friction) and \( F \sin 62^\circ = m*g = 117.6N \) (balances the weight). Solving these equations simultaneously will give the magnitude of the force \( F \).

Determine the force acting downwards

In part \( c \), the force \( F \) acts downwards at an angle \( 62^\circ \) from the horizontal. This decreases the normal force and, thus, the static friction as well. In this case, the horizontal and vertical components of \( F \) are also \( F \cos 62^\circ \) and \( -F \sin 62^\circ \) respectively. Thus, the normal force becomes \( F_n = m*g - F \sin 62^\circ \). Now, the static friction \( f_s = \mu_s * F_n \) and the horizontal component of \( F = f_s \). This equation can be solved to obtain the magnitude of \( F \) without causing the block to move.

Key concepts

Coefficient of Static Friction

Understanding the coefficient of static friction is crucial when dealing with objects at rest on surfaces. It's a dimensionless value that represents the frictional force between two static surfaces divided by the normal force exerted by one surface on the other. In our exercise, the coefficient of static friction, denoted as \( \mu_s \), plays a pivotal role in determining whether the steel block moves or stays put on the table.

The value of \( \mu_s = 0.52 \) tells us how much horizontal force is needed to overcome the frictional resistance before the block can start sliding. To better comprehend this coefficient, analyse it as a ratio: if the normal force is considered 1 unit, then the static frictional force equaled to the coefficient of static friction would be 0.52 units. Thus, the higher the coefficient, the more force is necessary to initiate motion,

which helps explain how different materials interact based on their surface characteristics. Imagine pushing a block across a slippery ice surface compared to a rough concrete one — the coefficient of static friction for ice would be much lower than for concrete, illustrating why less force is needed to move the block on ice.

Normal Force

The normal force is the perpendicular force exerted by a surface on any object resting upon it. This force is central to understanding many physical situations, including our textbook exercise. It's often denoted as \( F_n \) and is directly proportional to the object's weight which, in turn, depends on mass and gravity.

In the scenario provided, the normal force is calculated by multiplying the steel block's mass, \( 12 kg \), by the acceleration due to gravity, \( 9.8 m/s^2 \), resulting in a normal force of \( 117.6 N \).

Importance of the Normal Force in Static Friction

The normal force plays a key role in determining the maximum static frictional force because the static friction force is the product of this normal force and the coefficient of static friction. Therefore, any change in the normal force, such as adding a weight on top of the block or tilting the surface, will directly affect the force of static friction.

Static Friction Force Calculations

Calculating the static friction force is crucial not only in academic problems but also in real-world applications like engineering and safety designs. In our exercise, we calculated the static friction force to understand what horizontal force would initiate the block's movement.

The maximum static friction is found by multiplying the coefficient of static friction, \( \mu_s \), by the normal force, \( F_n \). Following our example, \( f_s = 0.52 \times 117.6N = 61.152N \). This means you would require a tad more than 61.152N of horizontal force to start moving the block.

However, different angles of applied forces also affect static friction. For instance, a force applied upward at an angle, like in part (b), can potentially reduce the normal force, due to the vertical component of the applied force counteracting the object's weight, which in turn affects the static friction force. Conversely, a force applied downward as in part (c) increases the normal force and accordingly the friction force as well.