Question

A crane of mass \(M=1.00 \cdot 10^{4} \mathrm{~kg}\) lifts a wrecking ball of mass \(m=1200 .\) kg directly upward. a) Find the magnitude of the normal force exerted on the crane by the ground while the wrecking ball is moving upward at a constant speed of \(v=1.00 \mathrm{~m} / \mathrm{s}\). b) Find the magnitude of the normal force if the wrecking ball's upward motion slows at a constant rate from its initial speed \(v=1.00 \mathrm{~m} / \mathrm{s}\) to a stop over a distance \(D=0.250 \mathrm{~m}\)

Short answer

Answer: a) The magnitude of the normal force exerted on the crane when the wrecking ball is moving upward with a constant speed is 1.14 x 10^5 N. b) The magnitude of the normal force exerted on the crane when the wrecking ball's upward motion slows down at a constant rate is 9.16 x 10^4 N.

Step-by-step solution

Given information

We are given: - The mass of the crane: \(M=1.00 \times 10^{4} \,\mathrm{kg}\) - The mass of the wrecking ball: \(m=1200\,\mathrm{kg}\) - The constant speed of the wrecking ball: \(v=1.00\,\mathrm{m/s}\) - The distance over which the wrecking ball slows down: \(D=0.250\,\mathrm{m}\)

Newton's second law

Newton's second law states that the acceleration of an object is equal to the net force acting on the object divided by the object's mass (\(F_{net} = ma\)). In this case, we will use this law to find the net forces acting on the crane and the wrecking ball.

Situation a: Constant speed of the wrecking ball

When the wrecking ball moves upward with a constant speed, its acceleration is zero (\(a=0\)). This means there are no net forces acting on the wrecking ball and crane system. When there are no net forces, we can conclude that the gravitational force on the system (\(F_{g,system} = (m+M)g\)) is equal to the normal force exerted by the ground (\(F_N\)): $$F_N = F_{g,system} = (m+M)g$$ We can now calculate the magnitude of the normal force:

Calculate the normal force (situation a)

Using the values given and standard gravity (\(g \approx 9.81 \,\mathrm{m/s^2}\)), we can find the normal force: $$F_N = (m+M)g = (1200\,\mathrm{kg} + 1.00 \times 10^{4} \,\mathrm{kg}) \times 9.81 \,\mathrm{m/s^2} = 1.14 \times 10^{5} \,\mathrm{N}$$ This is the magnitude of the normal force exerted on the crane by the ground while the wrecking ball is moving upward at a constant speed.

Situation b: Wrecking ball slows down at a constant rate

To find the magnitude of the normal force when the wrecking ball slows down, we first need to determine its acceleration. Let's use the following kinematic equation to find the final velocity (\(v_f\)) of the wrecking ball: $$v_f^2 = v_i^2 + 2aD$$ where \(v_i\) is the initial velocity of the wrecking ball, \(a\) is its acceleration, and \(D\) is the distance it moves. We know that the wrecking ball comes to a stop (\(v_f = 0\)), and we are given \(v_i = 1.00\,\mathrm{m/s}\) and \(D = 0.250\,\mathrm{m}\). We can solve for \(a\):

Calculate the acceleration (situation b)

Using the given values, we can find the acceleration: $$a = \frac{v_f^2 - v_i^2}{2D} = \frac{0^2 - (1.00\,\mathrm{m/s})^2}{2 \times 0.250\,\mathrm{m}} = -2\,\mathrm{m/s^2}$$ The negative sign indicates that the acceleration acts opposite to the motion, which confirms that the wrecking ball is slowing down.

Newton's second law (situation b)

Now we can use Newton's second law for the crane and wrecking ball system: $$F_{net} = (m+M)a = (1200\,\mathrm{kg} + 1.00 \times 10^{4} \,\mathrm{kg}) \times (-2\,\mathrm{m/s^2}) = -2.24 \times 10^{4} \,\mathrm{N}$$ We know that the net force is the difference between the normal force and the gravitational force \((F_N - F_{g,system})\). Therefore, we can now find the magnitude of the normal force:

Calculate the normal force (situation b)

Using the net force, we can find the normal force when the wrecking ball slows down: $$F_N = F_{g,system} + F_{net} = 1.14 \times 10^{5} \,\mathrm{N} - 2.24 \times 10^{4} \,\mathrm{N} = 9.16 \times 10^{4} \,\mathrm{N}$$ This is the magnitude of the normal force exerted on the crane by the ground when the wrecking ball's upward motion slows at a constant rate.

Key concepts

Newton's Second Law

Understanding Newton's second law is crucial when exploring the dynamics of motion. This law, often written as \( F_{net} = ma \), relates the net force (\( F_{net} \)) acting on an object to its mass (\( m \)) and acceleration (\( a \)). It's the backbone of classical mechanics and helps students analyze the forces in action, whether it's a car accelerating down a highway or a wrecking ball hoisted by a crane.

For instance, in our exercise, when the wrecking ball is lifted at a constant speed, its acceleration is zero. This leads to the conclusion that the net force is zero as well, meaning the forces acting on the system are in equilibrium. Therefore, the normal force exerted by the ground equals the gravitational force acting on the crane and the wrecking ball together. When the lifting speed changes, we then need to calculate the new acceleration to find the updated net force and, consequently, the new normal force the ground must exert.

Constant Velocity

A constant velocity indicates that an object is moving at a steady speed in a single direction. Importantly, 'constant velocity' also implies that there is no acceleration—since acceleration is the rate of change of velocity. When an object has a constant velocity, as cited in our wrecking ball example moving upward at \(1.00\,\mathrm{m/s}\), the forces acting upon it are balanced. This equilibrium means that the upward force (the tension in the crane's cable) directly opposes the downward force of gravity, resulting in a net force of zero.

Why Does Constant Velocity Equate to Zero Net Force?

In this state, Newton’s first law (often termed inertia) comes into play. It states that an object in motion at a constant velocity will stay in motion, unless acted upon by an external force. For our crane and wrecking ball, the absence of acceleration assures us that the net forces are indeed balanced, affording us the simplicity of calculating the normal force without the complication of unbalanced forces.

Acceleration Due to Gravity

The acceleration due to gravity, denoted by \(g\), is the acceleration that the Earth imparts to objects on or near its surface. Standard gravity is generally accepted to be \(9.81\,\mathrm{m/s^2}\) on Earth's surface. This constant acceleration affects any object under free fall and also contributes to the ‘weight’ (gravitational force) experienced by objects.

How Gravity Affects Our Calculations

Gravity plays a leading role in determining the normal force in our problem. Even when the crane holds the wrecking ball at a steady height or moves it at constant velocity, gravity imposes a continuous downward force equal to the mass of the crane and wrecking ball times \(g\). The normal force has to counteract this to maintain equilibrium, as it's the force exerted by a surface to support the weight of an object resting on it.

Kinematic Equations

Kinematic equations let us determine the motion characteristics of objects when acceleration is constant. The basic set includes formulas that relate displacement, initial and final velocities, acceleration, and time. These equations are powerful tools for analyzing motion, and they are particularly handy when figuring out problems involving objects accelerating under the influence of gravity.

In the situation where the wrecking ball comes to a stop over a certain distance, we use a kinematic equation to connect its initial velocity, the stopping distance, and the unknown acceleration. Knowing the final velocity is zero allows us to rearrange the equation to solve for the acceleration, which then feeds into Newton’s second law to determine the net force during this deceleration. This net force, when coupled with gravity’s pull, gives us insight into the change in the normal force during this phase of motion.