Question

A \(20.0-\mathrm{kg}\) box with a height of \(80.0 \mathrm{~cm}\) and a width of \(30.0 \mathrm{~cm}\) has a handle on the side that is \(50.0 \mathrm{~cm}\) above the ground. The box is at rest, and the coefficient of static friction between the box and the floor is \(0.280 .\) a) What is the minimum force, \(F,\) that can be applied to the handle so that the box will tip over without slipping? b) In what direction should this force be applied? b) In what direction should this force be applied?

Short answer

The minimum force required to tip the box without slipping is 98.1 N. b) In which direction should the force be applied to tip the box without causing it to slip? The force should be applied horizontally at the handle, acting perpendicular to the lever arm.

Step-by-step solution

Identify the forces acting on the box

We have three forces acting on the box: the gravitational force (\(F_g\)), the force applied to the handle (\(F\)), and the static friction force (\(F_{static}\)). We will consider the forces acting at the center of mass and the contact point between the box and the floor.

Calculate the gravitational force acting on the box

First, let's find the gravitational force (\(F_g\)) acting on the box. We can calculate this using the mass of the box (\(m\)) and the gravitational acceleration (\(g\)). The formula for the gravitational force is: \(F_g = m \times g\) Given the mass of the box is 20.0 kg and the gravitational acceleration is approximately 9.81 m/s², the gravitational force acting on the box is: \(F_g = 20.0\,\text{kg} \times 9.81\,\text{m/s}^2 = 196.2\,\text{N}\)

Calculate the force due to static friction

Next, we need to determine the force due to static friction between the box and the floor. This force will act horizontally and oppose the force applied to the handle. To find the maximum static friction force, we use the formula: \(F_{static} = \mu_s \times F_g\) Given the coefficient of static friction between the box and the floor is 0.280 and we already found the gravitational force applied to the box, the maximum static friction force is: \(F_{static} = 0.280 \times 196.2\,\text{N} = 54.94\,\text{N}\)

Calculate the torque due to the applied force

To find the minimum force required to tip the box, we need to balance the torque caused by the gravitational force with the torque caused by the applied force. The torque produced by the applied force (\(\tau_F\)) is given by: \(\tau_F = r \times F \times \sin{\theta}\) where \(r\) is the length of the lever arm, \(F\) is the applied force, and \(\theta\) is the angle between the lever arm and the force. We are given that the handle is 50.0 cm above the ground, and since the width of the box is 30.0 cm, the lever arm's length is 30.0 cm or 0.3 m. We want to apply the force in the direction perpendicular to the lever arm, so the angle \(\theta\) should be \(90^\circ\). So the torque due to applied force is: \(\tau_F = r \times F \times \sin{90^\circ} = 0.3\,\text{m} \times F\)

Calculate the torque due to the gravitational force

The torque produced by the gravitational force (\(\tau_g\)) acts at the center of mass of the box and has dimensions: \(\tau_g = \frac{L}{2} \times F_g \times \sin{\theta_g}\) where \(L\) is the distance between the center of mass and the pivot point, \(F_g\) is the gravitational force, and \(\theta_g\) is the angle between the lever arm and the gravitational force. The distance \(L\) is equal to half the width of the box, which is 15.0 cm or 0.15 m. The angle \(\theta_g\) is also \(90^\circ\). So the torque due to the gravitational force is: \(\tau_g = 0.15\,\text{m} \times 196.2\,\text{N} \times \sin{90^\circ} = 29.43\,\text{N}\cdot\text{m}\)

Set the torques equal and solve for the minimum force

To tip the box without slipping, we need to balance the torque due to the applied force with the torque due to the gravitational force. Therefore, we have: \(\tau_F = \tau_g\) \(0.3\,\text{m} \times F = 29.43\,\text{N}\cdot\text{m}\) Now, we can solve for the minimum force \(F\): \(F = \frac{29.43\,\text{N}\cdot\text{m}}{0.3\,\text{m}} = 98.1\,\text{N}\) a) The minimum force required to tip the box without slipping is 98.1 N.

Determine the direction of the applied force

b) To tip the box without slipping, the applied force must act perpendicular to the lever arm, which means it should be applied in the horizontal direction. Therefore, the force should be applied horizontally at the handle.