Question (52) Normal force on a girl sliding down a waterslide Wet n Wild has a waterslide called Daredevil Drop. At the steepest part of the slide, the angle is about 65. A girl goes down the slide. Her mass is 40 kg and the frictional force of the slide on her at its steepest angle is 20 N. (a) What is the normal (compression) component of the force by the slide on the girl? (b) What is the girl s acceleration? (c) What is the fundamental cause of the normal component of the force by the slide on the girl? Solution As always, begin by sketching a picture of the situation (see Figure 1). For simplicity, represent the girl as a box on the slide. Figure 1: A girl at the steepest part of a waterslide. (a) To solve the problem, apply the momentum principle. Draw a free-body diagram, showing the forces on the girl. There are really only two forces on the girl, (1) the force by the slide on the girl and (2) the gravitational force by Earth on the girl. However, the force by the slide can be resolved into two components a component perpendicular to the slide which is due to atomic compression of atoms in the slide and girl where they are in contact, and a component parallel to the slide which is friction. The free-body diagram for the girl is shown in Figure 2. Because the normal component and frictional component of the force by the slide on the girl are perpendicular and parallel to the slide, respectively, it s most convenient to define the coordinate system with +x parallel to the slide and +y perpendicular to the slide, as shown in Figure 3. But note that the gravitational force by Earth on the girl is NOT in the y-direction alone, but instead it has both an x component and a y-comonent as shown in Figure 4. Use right triangle trigonometry to find the x and y components of the gravitational force by Earth on the girl. Using geometry, you can shown that the angle between the y-component of the gravitational force and the vertical is the same as the angle of inclination of the slide. This right-triangle for the gravitational force and its components is shown in Figure 5. The hypotenuse of the right triangle is F grav. The x and y components of the gravitational force on the girl are found by examining the right-triangle.
Figure 2: A free-body diagram for the girl. Figure 3: The coordinate system. Figure 4: The x and y components of the gravitational force by Earth on the girl. F grav,y = F grav cos(70 ) = mg cos(70 ) = (40 kg)(9.8 N/kg) cos(70 ) = 134 N
Figure 5: The angle shown is the same as the angle of inclination of the slide. F grav,x = F grav sin(70 ) = mg sin(70 ) = (40 kg)(9.8 N/kg) sin(70 ) = 368 N Now, apply the momentum principle in component form as shown below. F net,x = p x = mv x = m v x = ma x and F net,y = p y = mv y = m v y = ma y Let s begin with the y-direction. The girl s y-velocity is zero and remains constant (and zero). Thus, the y-component of the net force is zero. F net,y = 0
Use the definition of net force as the sum of the forces and solve for the normal force. F net,y = 0 F N by slide on girl + F grav,y = 0 F N by slide on girl + 134 N = 0 F N by slide on girl = 134 N (b) To find the acceleration of the girl, apply the momentum principle in the x-direction. F net,x = ma x F grav,x + F fric by slide on girl = ma x 368 N + 20 N = ma x 348 N = (40 kg)a x a x = 348 N 40 kg a x = 8.7 m/s 2 Note that the acceleration is less than if the girl were freely falling (9.8 m/s 2 ). Yet, it s fairly high because of the steepness of the slide. Because the waterslide is curved, this acceleration will decrease as the slide flattens out. (c) The force by the slide on the girl is a contact force, meaning that it results from contact between the girl and slide. If they don t make contact, then they do not exert a force on each other. In other words, this is NOT a force that exists when the objects are separated, like the gravitational force or coulomb force. When the girl and slide make contact with each other, the atoms of the girl and the atoms of the slide are so close to each other, that they repel. This is a general property of non-bonding neutral atoms, when they get close to one another, they repel. As a result, the slide exerts a force on the girl that is perpendicular to the slide, which is called the normal force by the slide on the girl. In a ball-and-spring model, when the girl makes contact with the slide, the atoms (bonds) in the slide at the surface are compressed. The compression of the bonds (springs) exerts an opposite force on the girl that is perpendicular to the surface. For example, consider the ball-and-spring model of the a horizontal section of the slide shown in Figure 6. (Note: for simplicity, this is the horizontal part at the end of the slide.) When the girl sits on the slide, the bonds to the atoms near the surface of the slide compress as a result of interacting with the girl, as shown in the model in Figure 7. The net result of all of the compressed springs is a force on the girl that is perpendicular to the surface of the slide. Of course, the atomic balls and springs in the girl also compress and exert a normal force on the slide that is in the opposite direction as the normal force by the slide on the girl. This is consistent with Newton s third law (the Principle of Reciprocity).
Figure 6: A ball-and-spring model of the slide. Figure 7: A ball-and-spring model of the slide with the girl sitting on the slide.