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Step 3. Once a free-body diagram is drawn, Newton’s second law can be applied to solve the problem . This is done in [link] (d) for a particular situation. In general, once external forces are clearly identified in free-body diagrams, it should be a straightforward task to put them into equation form and solve for the unknown, as done in all previous examples. If the problem is one-dimensional—that is, if all forces are parallel—then they add like scalars. If the problem is two-dimensional, then it must be broken down into a pair of one-dimensional problems. This is done by projecting the force vectors onto a set of axes chosen for convenience. As seen in previous examples, the choice of axes can simplify the problem. For example, when an incline is involved, a set of axes with one axis parallel to the incline and one perpendicular to it is most convenient. It is almost always convenient to make one axis parallel to the direction of motion, if this is known.

Applying newton’s second law

Before you write net force equations, it is critical to determine whether the system is accelerating in a particular direction. If the acceleration is zero in a particular direction, then the net force is zero in that direction. Similarly, if the acceleration is nonzero in a particular direction, then the net force is described by the equation: F net = ma size 12{F rSub { size 8{ ital "net"} } = ital "ma"} {} .

For example, if the system is accelerating in the horizontal direction, but it is not accelerating in the vertical direction, then you will have the following conclusions:

F net x = ma size 12{F rSub { size 8{"net x"} } = ital "ma"} {} ,

F net y = 0 size 12{F rSub { size 8{"net y"} } =0} {} .

You will need this information in order to determine unknown forces acting in a system.

Step 4. As always, check the solution to see whether it is reasonable . In some cases, this is obvious. For example, it is reasonable to find that friction causes an object to slide down an incline more slowly than when no friction exists. In practice, intuition develops gradually through problem solving, and with experience it becomes progressively easier to judge whether an answer is reasonable. Another way to check your solution is to check the units. If you are solving for force and end up with units of m/s, then you have made a mistake.

Section summary

  • To solve problems involving Newton’s laws of motion, follow the procedure described:
    1. Draw a sketch of the problem.
    2. Identify known and unknown quantities, and identify the system of interest. Draw a free-body diagram, which is a sketch showing all of the forces acting on an object. The object is represented by a dot, and the forces are represented by vectors extending in different directions from the dot.
    3. Write Newton’s second law in the horizontal and vertical directions and add the forces acting on the object. If the object does not accelerate in a particular direction (for example, the x size 12{x} {} -direction) then F net x = 0 size 12{F rSub { size 8{"net x"} } =0} {} . If the object does accelerate in that direction, F net x = ma size 12{F rSub { size 8{"net x"} } = ital "ma"} {} .
    4. Check your answer. Is the answer reasonable? Are the units correct?

Problem exercises

A 5 . 00 × 10 5 -kg size 12{5 "." "00" times "10" rSup { size 8{5} } "- kg"} {} rocket is accelerating straight up. Its engines produce 1 . 250 × 10 7 N size 12{1 "." "250" times "10" rSup { size 8{7} } " N"} {} of thrust, and air resistance is 4 . 50 × 10 6 N size 12{4 "." "50" times "10" rSup { size 8{6} } " N"} {} . What is the rocket’s acceleration? Explicitly show how you follow the steps in the Problem-Solving Strategy for Newton’s laws of motion.

An object of mass m is shown. Three forces acting on it are tension T, shown by an arrow acting vertically upward, and friction f and gravity m g, shown by two arrows acting vertically downward.

Using the free-body diagram:

F net = T f m g = ma size 12{F rSub { size 8{"net"} } =T - f= ital "ma"} {} ,

so that

a = T f mg m = 1 . 250 × 10 7 N 4.50 × 10 6 N ( 5.00 × 10 5 kg ) ( 9. 80 m/s 2 ) 5.00 × 10 5 kg = 6.20 m/s 2 size 12{a= { {T` - `f` - ` ital "mg"} over {m} } = { {1 "." "250" times "10" rSup { size 8{7} } " N" - 4 "." "50" times "10" rSup { size 8{"6 "} } N - \( 5 "." "00" times "10" rSup { size 8{5} } " kg" \) \( 9 "." "80 m/s" rSup { size 8{2} } \) } over {5 "." "00" times "10" rSup { size 8{5} } " kg"} } ="6" "." 20" m/s" rSup { size 8{2} } } {} .

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Source:  OpenStax, Abe advanced level physics. OpenStax CNX. Jul 11, 2013 Download for free at http://legacy.cnx.org/content/col11534/1.3
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