Weight Normal Force Weight The force of gravity
- Slides: 18
Weight & Normal Force • Weight The force of gravity on an object. • Write as FG W. • Consider an object in free fall. Newton’s 2 nd Law: ∑F = ma • If no other forces are acting, only FG ( W) acts (in the vertical direction). ∑Fy = may Or: (down, of course) • SI Units: Newtons (just like any force!). g = 9. 8 m/s 2 If m = 1 kg, W = 9. 8 N
Normal Force • Suppose an object is at rest on a table. No motion, but does the force of gravity stop? OF COURSE NOT! • But, the object does not move: Newton’s 2 nd Law is ∑F = ma = 0 So, there must be some other force acting besides gravity (weight) to have ∑F = 0. • That force The Normal Force FN (= n) “Normal” is a math term for perpendicular ( ) FN is to the surface & equal & opposite to the weight (true in this simple case only!) CAUTION!! FN isn’t always = & opposite to the weight, as we’ll see!
Example “Free Body Diagram” for the monitor. Shows all forces on it, in proper directions. Monitor at rest on table. Force of monitor on table ≡ Fmt. Force of table on monitor ≡ Ftm keeps monitor from falling. Ftm & Fmt are 3 rd Law action-reaction pairs. Forces on monitor are “Normal Force” n & weight Fg. 2 nd Law for monitor in vertical direction: ∑Fy = 0 = n - Fg. So, n = Fg = mg. So n = mg. They are equal & in opposite directions, BUT THEY ARE NOT action-reaction pairs (they act on the SAME object, not on different objects!)
Normal Force Where does the normal force come from?
Normal Force Where does the normal force come from? From the other object!!!
Normal Force Where does the normal force come from? From the other object!!! Is the normal force ALWAYS equal & opposite to the weight?
Normal Force Where does the normal force come from? From the other object!!! Is the normal force ALWAYS equal & opposite to the weight? NO!!!
An object at rest must have no net force on it. If it is sitting on a table, the force of gravity is still there; what other force is there? The force exerted perpendicular to a surface is called the Normal Force FN. It is “Free Body Diagrams” for Lincoln. Show all forces in proper directions. exactly as large as needed to balance the force from the object. (If the required force gets too big, something breaks!) Newton’s 2 nd Law for Lincoln: ∑F = ma = 0 or FN – FG = 0 or FN = FG = mg FN & FG AREN’T action-reaction pairs from N’s 3 rd Law! They’re equal & opposite because of N’s 2 nd Law! FN & FN ARE action-reaction pairs!!
Example m = 10 kg l Find: The Normal force on the box from the table for Figs. a. , b. , c.
Example m = 10 kg The normal force is NOT l always equal & opposite to the weight!! Find: The Normal force on the box from the table for Figs. a. , b. , c. Always use N’s 2 nd Law to CALCULATE l F N!
Example m = 10 kg The normal force is NOT always equal & opposite to the weight!! l Find: The Normal force on the box from the table for Figs. a. , b. , c. Always use N’s 2 nd Law to CALCULATE F N!
Example m = 10 kg The normal force is NOT always equal & opposite to the weight!! Find: The Normal force on the box from the table for Figs. a. , b. , c. Always use N’s 2 nd Law to CALCULATE F N!
Example l What happens when a m = 10 kg, ∑F = ma person pulls upward on F – mg = ma P l the box in the previous 100 – 89 = 10 a example with a force a = 0. 2 m/s 2 m = 10 kg greater than the box’s ∑F = ma weight, say 100. 0 N? l The box will accelerate upward because FP – mg = ma FP > mg!! Note: The normal force is zero here l because the mass isn’t in contact with a surface!
Example What happens when a m = 10 kg, ∑F = ma person pulls upward on FP – mg = ma the box in the previous 100 – 89 = 10 a example with a force a = 0. 2 m/s 2 m = 10 kg greater than the box’s ∑F = ma weight, say 100. 0 N? The box will accelerate upward because FP – mg = ma FP > mg!! Note The normal force is zero here because the mass isn’t in contact with a surface!
Example: Apparent “weight loss” A 65 -kg (mg = 640 N) woman descends in an elevator that accelerates at a rate a = 0. 20 g downward. She stands on a scale that reads in kg. (a) During this acceleration, what is her weight & what does the scale read? (b) What does the scale read when the elevator descends at a constant speed of 2. 0 m/s?
Example: Apparent “weight loss” A 65 -kg (mg = 640 N) woman descends in an elevator that accelerates at a rate a = 0. 20 g downward. She stands on a scale that reads in kg. (a) During this acceleration, what is her weight & what does the scale read? (b) What does the scale read when the elevator descends at a constant speed of 2. 0 m/s? Reasoning to get the solution using from Newton’s Laws To use Newton’s 2 nd Law for her, ONLY the forces acting on her are included. By Newton’s 3 rd Law, the normal force FN acting upward on her is equal & opposite to the scale reading. So, the numerical value of FN is equal to the “weight” she reads on the scale! Obviously, FN here is NOT equal & opposite to her true weight mg!! How do we find FN? As always, WE APPLY NEWTON’S 2 ND LAW TO HER!!
Example: Apparent “weight loss” A 65 -kg (mg = 637 N) woman descends in an elevator that accelerates at a rate a = 0. 20 g downward. She stands on a scale that reads in kg. (a) During this acceleration, what is her weight & what does the scale read? (b) What does the scale read when the elevator descends at a constant speed of 2. 0 m/s? Solution (a) Newton’s 2 nd Law applied to the woman is (let down be positive!): ∑F = ma Since a is a 1 d vector pointing down, this gives: mg – FN = ma so FN = mg - ma = m(g – 0. 2 g) = 0. 8 mg which is numerically equal to the scale reading by Newton’s 3 rd Law!! So if she trusts the scale (& if she doesn’t know N’s Laws!), she’ll think that she has lost 20% of her body weight!!
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