Normal, Tension, and Other Examples of Forces
Learning Objectives
By the end of this section, you will be able to: Define normal and tension forces.
 Apply Newton's laws of motion to solve problems involving a variety of forces.
 Use trigonometric identities to resolve weight into components.
Normal Force
Common Misconception: Normal Force (N) vs. Newton (N)
Example 1. Weight on an Incline, a TwoDimensional Problem
Strategy
This is a twodimensional problem, since the forces on the skier (the system of interest) are not parallel. The approach we have used in twodimensional kinematics also works very well here. Choose a convenient coordinate system and project the vectors onto its axes, creating two connected onedimensional problems to solve. The most convenient coordinate system for motion on an incline is one that has one coordinate parallel to the slope and one perpendicular to the slope. (Remember that motions along mutually perpendicular axes are independent.) We use the symbols ⊥ and ∥ to represent perpendicular and parallel, respectively. This choice of axes simplifies this type of problem, because there is no motion perpendicular to the slope and because friction is always parallel to the surface between two objects. The only external forces acting on the system are the skier’s weight, friction, and the support of the slope, respectively labeled w, f, and N in Figure 2. N is always perpendicular to the slope, and f is parallel to it. But w is not in the direction of either axis, and so the first step we take is to project it into components along the chosen axes, defining w_{∥} to be the component of weight parallel to the slope and w_{⊥} the component of weight perpendicular to the slope. Once this is done, we can consider the two separate problems of forces parallel to the slope and forces perpendicular to the slope.Solution
The magnitude of the component of the weight parallel to the slope is(a) Neglecting friction. Since the acceleration is parallel to the slope, we need only consider forces parallel to the slope. (Forces perpendicular to the slope add to zero, since there is no acceleration in that direction.) The forces parallel to the slope are the amount of the skier’s weight parallel to the slope w_{∥} and friction f. Using Newton’s second law, with subscripts to denote quantities parallel to the slope,
(b) Including friction. We now have a given value for friction, and we know its direction is parallel to the slope and it opposes motion between surfaces in contact. So the net external force is now
Discussion
Since friction always opposes motion between surfaces, the acceleration is smaller when there is friction than when there is none. In fact, it is a general result that if friction on an incline is negligible, then the acceleration down the incline is a = g sin θ, regardless of mass. This is related to the previously discussed fact that all objects fall with the same acceleration in the absence of air resistance. Similarly, all objects, regardless of mass, slide down a frictionless incline with the same acceleration (if the angle is the same).Resolving Weight into Components
TakeHome Experiment: Force Parallel
Tension
T = w = mg.
For a 5.00kg mass, then (neglecting the mass of the rope) we see thatT = mg = (5.00 kg)(9.80 m/s^{2}) = 49.0 N
If we cut the rope and insert a spring, the spring would extend a length corresponding to a force of 49.0 N, providing a direct observation and measure of the tension force in the rope. Flexible connectors are often used to transmit forces around corners, such as in a hospital traction system, a finger joint, or a bicycle brake cable. If there is no friction, the tension is transmitted undiminished. Only its direction changes, and it is always parallel to the flexible connector. This is illustrated in Figure 5 (a) and (b).Example 2. What Is the Tension in a Tightrope?
Strategy
As you can see in the figure, the wire is not perfectly horizontal (it cannot be!), but is bent under the person’s weight. Thus, the tension on either side of the person has an upward component that can support his weight. As usual, forces are vectors represented pictorially by arrows having the same directions as the forces and lengths proportional to their magnitudes. The system is the tightrope walker, and the only external forces acting on him are his weight w and the two tensions T_{L} (left tension) and T_{R} (right tension), as illustrated. It is reasonable to neglect the weight of the wire itself. The net external force is zero since the system is stationary. A little trigonometry can now be used to find the tensions. One conclusion is possible at the outset—we can see from part (b) of the figure that the magnitudes of the tensions T_{L} and T_{R} must be equal. This is because there is no horizontal acceleration in the rope, and the only forces acting to the left and right are T_{L} and T_{R}. Thus, the magnitude of those forces must be equal so that they cancel each other out.Whenever we have twodimensional vector problems in which no two vectors are parallel, the easiest method of solution is to pick a convenient coordinate system and project the vectors onto its axes. In this case the best coordinate system has one axis horizontal and the other vertical. We call the horizontal the xaxis and the vertical the yaxis.
Solution
First, we need to resolve the tension vectors into their horizontal and vertical components. It helps to draw a new freebody diagram showing all of the horizontal and vertical components of each force acting on the system.F_{netx} = T_{Lx} − T_{Rx}.
Equating T_{L x} and T_{Rx}:
Thus,
Discussion
Note that the vertical tension in the wire acts as a normal force that supports the weight of the tightrope walker. The tension is almost six times the 686N weight of the tightrope walker. Since the wire is nearly horizontal, the vertical component of its tension is only a small fraction of the tension in the wire. The large horizontal components are in opposite directions and cancel, and so most of the tension in the wire is not used to support the weight of the tightrope walker.Extended Topic: Real Forces and Inertial Frames
Earth’s rotation is slow enough that Earth is nearly an inertial frame. You ordinarily must perform precise experiments to observe fictitious forces and the slight departures from Newton’s laws, such as the effect just described. On the large scale, such as for the rotation of weather systems and ocean currents, the effects can be easily observed.
The crucial factor in determining whether a frame of reference is inertial is whether it accelerates or rotates relative to a known inertial frame. Unless stated otherwise, all phenomena discussed in this text are considered in inertial frames.
All the forces discussed in this section are real forces, but there are a number of other real forces, such as lift and thrust, that are not discussed in this section. They are more specialized, and it is not necessary to discuss every type of force. It is natural, however, to ask where the basic simplicity we seek to find in physics is in the long list of forces. Are some more basic than others? Are some different manifestations of the same underlying force? The answer to both questions is yes, as will be seen in the next (extended) section and in the treatment of modern physics later in the text.
PhET Explorations: Forces in 1 Dimension
Section Summary
 When objects rest on a surface, the surface applies a force to the object that supports the weight of the object. This supporting force acts perpendicular to and away from the surface. It is called a normal force, N.
 When objects rest on a nonaccelerating horizontal surface, the magnitude of the normal force is equal to the weight of the object:
N = mg  When objects rest on an inclined plane that makes an angle θ with the horizontal surface, the weight of the object can be resolved into components that act perpendicular (${\mathbf{\text{w}}}_{\perp}$) and parallel (${\mathbf{\text{w}}}_{\parallel}$) to the surface of the plane. These components can be calculated using:
 ${w}_{\parallel}=w\sin({\theta})=mg\sin({\theta})$

${w}_{\perp}=w\cos({\theta})=mg\cos({\theta})$
 The pulling force that acts along a stretched flexible connector, such as a rope or cable, is called tension, T. When a rope supports the weight of an object that is at rest, the tension in the rope is equal to the weight of the object:
T = mg.  In any inertial frame of reference (one that is not accelerated or rotated), Newton’s laws have the simple forms given in this chapter and all forces are real forces having a physical origin.
Conceptual Questions
Problems & Exercises
Glossary
 inertial frame of reference:
 a coordinate system that is not accelerating; all forces acting in an inertial frame of reference are real forces, as opposed to fictitious forces that are observed due to an accelerating frame of reference
 normal force:
 the force that a surface applies to an object to support the weight of the object; acts perpendicular to the surface on which the object rests
 tension:
 the pulling force that acts along a medium, especially a stretched flexible connector, such as a rope or cable; when a rope supports the weight of an object, the force on the object due to the rope is called a tension force
Selected Solutions to Problems & Exercises
1. (a) 0.11 m/s^{2} (b) 1.2 × 10^{4} N3. (a) 7.84 × 10^{4} N (b) 1.89 × 10^{3} N. This is 2.41 times the tension in the vertical strand.
5. Newton’s second law applied in vertical direction gives