Newton’s 2 nd Law some examples

Newton’s 2nd Lawsome examples

More examples… Xavier, who was stopped at a light, accelerated forward at 6 m/s2 when the light changed. He had dice hanging from his rear view mirror. What angle did they make with the vertical during this acceleration? 1- draw free body diagram 2- write Newton’s 2nd law for BOTH directions SFx = max Tsinq = max Tcosq = mg tanq = a/g q = 31.47 degrees q T SFy = may Tcosq – mg = may = 0 mg

An “Atwood’s Machine” Two masses of 5 kg and 2 kg are suspended from a massless, frictionless pulley, When released from rest, what is their acceleration? What is the Tension in the string? 1- Draw a free body diagram 2- Write Newton’s Second Law for EXTERNAL forces acting on the whole system. SFext = mtotala The Tension in the string is an “internal” force. The only “external” forces are the gravitational forces of weight, which oppose each other. So… SFext = m1g – m2g = mtotala Solving for the acceleration yields a = 4.2 m/s2 Now, to find the Tension, we must “zoom in” on mass 2: SF= T – m2g = m2a T = m2a + m2g T = 28 N a +

Objects on Inclines- sliding down, no friction The free-body diagram ALWAYS comes first: Draw the weight vector, mg Draw the Normal force vector. Are there any other forces??? Since the motion is parallel to the plane, ROTATE the axis from horizontal and vertical to “parallel” and “perpendicular”. Then draw the components of the weight vector, both perpendicular and parallel to the incline. The components help form a right triangle. Label the angle. Does the Normal force balance with the force of weight, mg? NO! What force must balance the Normal force? N = mgcosq N Does all of the weight, mg, pull the box down the incline? Write Newton’s Second Law for the forces on box parallel to the plane with “down” being negative. S F = ma -mgsin q = ma No! Only mgsinq q mg q

Inclines: pushed upward, no friction Draw the weight vector, mg Draw the Normal force vector. Draw the components of the weight vector, both perpendicular and parallel to the incline. The components help form a right triangle. Label the angle. Write Newton’s Second Law for the box. S F = ma FA - mgsin q = ma N FA q mg q

Inclines: pushed downward, no friction Draw the weight vector, mg Draw the Normal force vector. Draw the components of the weight vector, both perpendicular and parallel to the incline. The components help form a right triangle. Label the angle. Write Newton’s Second Law for the box. S F = ma - FA - mgsin q = ma N FA q mg q

Spring Force If a mass is suspended from a spring, two forces act on the mass: its weight and the spring force. The spring force for many springs is given by Fs = kx, Where x is the distance the spring is stretched (or compressed) from its normal length and “k” is called the spring constant, which tells the stiffness of the spring. The stiffer the spring, the larger the spring constant k. This relationship is known as “Hooke’s Law”. Fs mg

Spring Force If the mass is at rest, then the two forces are balanced, so that kx = mg These balanced forces provide a quick way to determine the spring constant of a spring, or anything springy, like a rubber band: Hang a known mass from the spring and measure how much it stretches, then solve for “k” ! k = mg/x Fs= kx mg

If the box is not accelerating up or down, then the support force from the table, the “Normal force”, must by equal to the Weight Normal force, N Weight = mg

But… what if you were accelerating up or down? The Normal force would NOT be equal to your weight if you’re accelerating up or down. And… you feel heavier or lighter! This feeling is called your “apparent weight”.

Apparent Weight When you ride an elevator, you “feel” heavier or lighter than you actually weigh because of the acceleration of the elevator. Your own sense of your weight is called your “apparent weight”. That sensation comes from the support underneath us. If the floor underneath you quickly starts pushing up hard upon your feet, you feel heavier. If the floor underneath your feet quickly starts moving downward, you feel lighter. If the floor underneath your feet falls away completely, you feel weightless!

N mg A 50 kg woman steps on a scale in an elevator that accelerates upward at 1.5 m/s2. What is her weight (use g = 10 m/s2) ? Weight = mg = 500 N How heavy does she feel- in other words, what is her APPARENT weight? SF = ma N – mg = ma Her APPARENT weight is what she feels like she weighs, which is determined by how hard the floor is pushing up against her- the “Normal” force. N = mg + ma N = 500 N + 50 kg x 1.5 m/s2 Apparent weight = 575 N

Do free-falling objects REALLY accelerate toward the Earth at 9.8 m/s2? No, because of air resistance. Air resistance is a force that pushes up on an object as it falls. The faster you fall… The greater the air resistance. Eventually, the air resistance pushing up on you is just as large as your weight that is pulling down on you!!

The faster the man falls, the more air resistance pushes up on him. Eventually, there will be just as much air resistance pushing up on him as his weight pulling him down. What will be the NET force acting then? What will be his acceleration?

Once the air resistance pushing up is as large as the weight pushing down, the NET force acting on you is ZERO! If the net force is zero, what is your acceleration? ZERO! This doesn’t mean you stop in mid air. But it does mean that you stop accelerating! You still continue to fall towards the Earth, but you don’t pick up any more velocity- you continue to fall towards the Earth at the same velocity. This speed is called your “terminal velocity” You will reach your terminal velocity when Air resistance = your weight

Which one will have a faster terminal velocity? You don’t reach terminal velocity until the air resistance grows to as large a force as your weight. The more massive skydiver will have a faster terminal velocity and hit the ground at a faster speed

Newton’s 2 nd Law some examples