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A person is holding a bag of dog food at some height from a table. He is exerting a force F sub hand, shown by a vector arrow in upward direction, and the weight W of the bag is acting downward, shown by a vector arrow having the same length as vector F sub hand. In a free-body diagram two forces are acting on the red point; one is F sub hand, shown by a vector arrow upward, and another is the weight W, shown by a vector arrow having the same length as vector F sub hand but pointing downward. (b) The bag of dog food is on the table, which deforms due to the weight W, shown by a vector arrow downward; the normal force N is shown by a vector arrow pointing upward having the same length as W. In the free-body diagram, vector W is shown by an arrow downward and vector N is shown by an arrow having the same length as vector W but pointing upward.
(a) The person holding the bag of dog food must supply an upward force F hand size 12{F rSub { size 8{"hand"} } } {} equal in magnitude and opposite in direction to the weight of the food w size 12{w} {} . (b) The card table sags when the dog food is placed on it, much like a stiff trampoline. Elastic restoring forces in the table grow as it sags until they supply a force N size 12{N} {} equal in magnitude and opposite in direction to the weight of the load.

We must conclude that whatever supports a load, be it animate or not, must supply an upward force equal to the weight of the load, as we assumed in a few of the previous examples. If the force supporting a load is perpendicular to the surface of contact between the load and its support, this force is defined to be a normal force     and here is given the symbol N size 12{N} {} . (This is not the unit for force N.) The word normal means perpendicular to a surface. The normal force can be less than the object’s weight if the object is on an incline, as you will see in the next example.

Common misconception: normal force (n) vs. newton (n)

In this section we have introduced the quantity normal force, which is represented by the variable N size 12{N} {} . This should not be confused with the symbol for the newton, which is also represented by the letter N. These symbols are particularly important to distinguish because the units of a normal force ( N size 12{N} {} ) happen to be newtons (N). For example, the normal force N size 12{N} {} that the floor exerts on a chair might be N = 100 N size 12{N="100"" N"} {} . One important difference is that normal force is a vector, while the newton is simply a unit. Be careful not to confuse these letters in your calculations! You will encounter more similarities among variables and units as you proceed in physics. Another example of this is the quantity work ( W size 12{W} {} ) and the unit watts (W).

Weight on an incline, a two-dimensional problem

Consider the skier on a slope shown in [link] . Her mass including equipment is 60.0 kg. (a) What is her acceleration if friction is negligible? (b) What is her acceleration if friction is known to be 45.0 N?

A skier is skiing down the slope and the slope makes a twenty-five degree angle with the horizontal. Her weight W, shown by a vector vertically downward, breaks into two components—one is W parallel, which is shown by a vector arrow parallel to the slope, and the other is W perpendicular, shown by a vector arrow perpendicular to the slope in the downward direction. Vector N is represented by an arrow pointing upward and perpendicular to the slope, having the same length as W perpendicular. Friction vector f is represented by an arrow along the slope in the uphill direction. IIn a free-body diagram, the vector arrow W for weight is acting downward, the vector arrow for f is shown along the direction of the slope, and the vector arrow for N is shown perpendicular to the slope.
Since motion and friction are parallel to the slope, it is most convenient to project all forces onto a coordinate system where one axis is parallel to the slope and the other is perpendicular (axes shown to left of skier). N size 12{N} {} is perpendicular to the slope and f is parallel to the slope, but w size 12{w} {} has components along both axes, namely w size 12{w rSub { size 8{ ortho } } } {} and w . N size 12{N} {} is equal in magnitude to w size 12{w rSub { size 8{ ortho } } } {} , so that there is no motion perpendicular to the slope, but f size 12{f} {} is less than w size 12{w rSub { size 8{ \lline \lline } } } {} , so that there is a downslope acceleration (along the parallel axis).

Strategy

This is a two-dimensional problem, since the forces on the skier (the system of interest) are not parallel. The approach we have used in two-dimensional kinematics also works very well here. Choose a convenient coordinate system and project the vectors onto its axes, creating two connected one -dimensional 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 size 12{ ortho } {} and size 12{ \lline \lline } {} 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 size 12{w} {} , f size 12{f} {} , and N size 12{N} {} in [link] . N size 12{N} {} is always perpendicular to the slope, and f size 12{f} {} is parallel to it. But w size 12{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 size 12{w rSub { size 8{ \lline \lline } } } {} to be the component of weight parallel to the slope and w size 12{w rSub { size 8{ ortho } } } {} 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.

Practice Key Terms 3

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Source:  OpenStax, College physics for ap® courses. OpenStax CNX. Nov 04, 2016 Download for free at https://legacy.cnx.org/content/col11844/1.14
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