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## Newton’s Laws of Motion

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**Basic Concepts**• Force – push or pull on an object - Vector quantity • Mass – amount of matter in a body. - Measured with a balance - Scalar quantity HFinks '07**First Law**• Also known as the law of inertia • An object will remain at rest or continue to move at a constant speed unless acted upon by a net force. • Net force – vector sum of all forces acting on an object. HFinks '07**Second Law**• When a net force acts on a body, the acceleration is directly proportional to the force and indirectly proportional to the mass. • Equation: ∑F = ma or Fnet = ma Net force on an object = mass x acceleration Net = sum of the forces • Units of measurement for force metric system - Newton (N) metric system - Dyne (dyn) english system - Pound (lb) HFinks '07**Third Law**For every action, there is an equal and opposite reaction. Forces occur in pairs. Net force and acceleration will always be in the same direction. HFinks '07**Examples and Labeling**of Forces HFinks '07**Applied Force (Fa)**• Pushing, pulling or lifting an object. • Applied force is a vector quantity. HFinks '07**Applied Force (Fa)**Example 1: Lifting an object Fa HFinks '07**Applied Force (Fa)**Example 2: Pulling an Object Fa HFinks '07**Applied Force (Fa)**Example 3: Pushing an Object Fa HFinks '07**Applied Force (Fa)**Example 4: Pulling an object up an inclined plane Fa HFinks '07**Normal Force (FN)**• Normal force, FN , is a component of the force the surface exerts on an object. • Normal force is always perpendicular to the surface. • Normal force is a vector quantity. HFinks '07**Normal Force (FN)**Example 1: Block on floor FN HFinks '07**Normal Force (FN)**Example 2: Block on wall FN HFinks '07**Normal Force (FN)**Example 3: Block on inclined plane FN HFinks '07**Weight (W)**Weight – depends on the acceleration due to gravity. - Measured with a spring scale. - Vector quantity - Is a force - Equation: W = mg - Units of measurement metric system - Newton metric system - Dyne english system - Pound HFinks '07**Comparing Units**F = ma W = mg = kg m/s2 = kg m/s2 = N = N N = Newton HFinks '07**Weight (W) and Mass (m)**• Weight and mass are not the same. • Weight changes with a change in acceleration due to gravity. • Mass does not change with a change in acceleration due to gravity. Example: • Your weight will change if you go from the earth to the moon. • Your mass will remain the same. HFinks '07**Weight (W)**Example 1: Block on floor W HFinks '07**Weight (W)**Example 2: Block on wall W HFinks '07**Weight (W)**• Example 3: Block on inclined plane W HFinks '07**Friction**• Occurs when objects are in contact • Acts opposite to the direction of motion • Two kinds of frictional forces Static (object at rest) = fs Kinetic (object is moving) = fk • Equations Static fsmax =µsFN Kinetic fk = µkFN HFinks '07**Coefficient of Friction (µ)**• Ratio of the frictional force to the normal force. That is, you are comparing one force to another. StaticKinetic (sliding) µs = fs µk = fk FN FN • No unit of measurement because you are dividing a force by a force. HFinks '07**Kinetic Frictional Force (fk)**• The frictional force always act opposite to the applied force. Think of these two forces as being x-components. One is positive and the other is negative. Fa fk Direction of motion HFinks '07**Kinetic Frictional Force (fk)**Example 1: Box pulled across a floor Fa fk HFinks '07**Kinetic Frictional Force (fk)**Example 2: Object pulled up an inclined plane Fa fk HFinks '07**Kinetic Frictional Force (fk)**• Example 3: Object sliding down an inclined plane • No applied force fk HFinks '07**Static Frictional Force (fs)**- Friction between atoms or molecules - No applied force HFinks '07**Frictional Force (fs)**Example 1: Box at rest on a floor fs HFinks '07**Frictional Force (fs)**• Example 2: Object at rest on an inclined plane fs HFinks '07**Fnet = ma**Applications of this equation in the x direction Applied Force (Fa) and Frictional Force (fk) HFinks '07**Fnet = ma**Applications of this equation in the y direction Normal Force (FN) and Weight (W) Applied Force (FN) and Weight (W) HFinks '07**Applied Force (Fa) and Weight (W)**Example 1: Lifting Fa W Fa – W = ma HFinks '07**Normal Force (FN) and Weight (W)**Example 2: Block on floor FN W FN - W = ma HFinks '07**Solving Force Problems**• Label all forces • Write an expression for the forces acting in the x direction. • Write an expression for the forces acting in the y direction. • Substitute the known values into the expressions. • Solve for the unknown. HFinks '07**Example 1**What is the magnitude and direction of FN? FN • Label forces • (No forces in x direction) b. Write expression Mass = 20.0 kg FN - W = ma At rest ay = 0 m/s2 W = mg = 196 N c. FN - W = ma FN = W + ma =196 N + (20.0 kg)(0 m/s2) FN = 196 N, upward HFinks '07**Example 2**A 12.0 kg object is pulled upward by a massless rope with an acceleration of 3.00 m/s2. What is the tension (T) in the rope? Fa = T a = 3.00 m/s2 m = 12.0 kg W = (12.0 kg)(9.80 m/s2) = 118 N T – W = ma T = 118 N + (12.0 kg)(3.00 m/s2) T = W + ma T = 154 N HFinks '07**Example 3**A 12.0 kg object is accelerating downward at 3.00 m/s2. What is the tension (T) in the rope? Fa = T a = -3.00 m/s2 m = 12.0 kg W = (12.0 kg)(9.80 m/s2) = 118 N T – W = ma T = 118 N + (12.0 kg)(-3.00 m/s2) T = W + ma T = 82.0 N HFinks '07**Example 4**A 100. N crate is pulled across the floor. The tension in the rope is 75.0 N. Calculate the maximum frictional force. (µk = 0.5) a FN a. Label forces Fa = T fk b. Write expressions W = 100 n Fk = µk FN = (0.5)(100 N) Fk = 50.0 N HFinks '07**Example 4**Calculate the resultant force (Fnet) in Example 4. a a. Label forces FN=100. N Fa = T = 75.0 N Fk =50.0 N b. Write expressions W = 100 n Fy = FN – W Fx = T – fk FN = W = 75.0 N – 50.0 N Fy = 100. N – 100. N Fx = 25.0 N Fy = 0 N HFinks '07**Example 4**Calculate the acceleration of the object in Example 4. a FN a. Label forces fk= 50.0 N b. Write expressions for direction of motion only. Fa = T= 75.0 N W = 100. N m = w = 100. N = 10.2 kg g 9.80 m/s2 T – fk = ma a = 75.0 N – 50.0 N a = T – fk10.2 kg m a = 2.45 m/s2 HFinks '07**Inclined Plane**HFinks '07**Example 5**A 20.0 kg object is at rest on an inclined plane (µs = 0.10) What is the value of fs? Ө = 10.0º fs = µsFN FN ≠ W Reason: They aren’t opposite each other “W” is a vector quantity and is acting at an angle. Calculate x- and y-components, FN fs W Ө Next HFinks '07**Example 5**W = mg = 196 N Wy = W cos Ө Wx = W sin Ө Note: Ө is the same for both the large and small right triangles. Wx will always act “down” the incline.. ( -Wx) FN Wy fs Ө W Wx Ө Next HFinks '07**Example 5**FN = Wy Fs = µsFN = µsWy = µs ( W cos Ө) = (0.10)(196 N)(cos 10.0º) fs = 19.3 N FN Wy fs Ө W Wx Ө HFinks '07**Example 6**A 20.0 kg object is sliding down an incline at a constant velocity under the influence of gravity. Find the value of fK. fk HFinks '07**Example 6**Label Forces and write expressions Object isn’t moving in the y direction. So…..don’t use the y expression: FN - Wy = ma Object is moving along the incline. Use…. ∑Fx fk - Wx = ma a = 0. Object is moving at a constant velocity. So….. Fk = Wx FN fk Wy Ө W Wx Ө Next HFinks '07**Example 6**fk = Wx = W sin Ө =196N sin 10.0º fk = 34.0 N FN fk Wy Ө W Wx Ө Next HFinks '07**Example 7**A 20.0 kg object is pulled up a 10.0º incline and µk = 0.05. The object is accelerated at a rate of 1.25 m/s2. What is the value of the applied force? W = 196 N (i) Fa - fk - Wx = ma Fa = fk + Wx + ma *’a” is positive. The object is moving up the incline. fk = µkFN = µkWy fk = µkW cos Ө FN Fa Wy fk Ө W Wx Ө Next HFinks '07**Example 7**W = 196 N • (i) Fa - fk - Wx = ma • Fa = fk + Wx + ma • Fa = (µkW cos Ө) +W sin Ө + ma • = (0.05)(196 N)(cos 10.0º) + (196 N)(sin 10.0º) + (20.0 kg)(1.25 m/s2) • = 9.65 N + 34.0 N + 25.0 N • Fa = 68.7 N FN Fa Wy fk Ө W Wx Ө HFinks '07