Newtons 2nd Law in 2D Forces and kinematics

Forces on a stationary object What forces are acting on this stationary object? Fnormal mg

Forces on a non-stationary object If the same object is positioned on an inclined plane, how does that affect to the two forces shown in the previous slide? The force due to gravity does not change direction, but the normal force acts perpendicular to the surface it’s on. Fnormal mg ao

Forces on a non-stationary object What other forces now apply, and why? The resultant force cause by gravity which moves the object down the inclined plane can be broken into X and Y components as shown. Fnormal mg sin(a) mg cos(a) mg ao

Forces on a non-stationary object If the ball has a mass of 5kg, the angle is 10o, and its position is 2 meters up the ramp, then what will be its velocity when it reaches the bottom? First find the acceleration: Fdown = 50sin(10) = 5a a = 1.74m/s2 Then use the other known values to solve: s = 2m u = 0 v = 2.64m/s Fnormal mg sin(a) mg cos(a) mg ao

Forces on a non-stationary object A common follow up question on an exam may be something like, if upon reaching the ground the object experiences a constant frictional resistance of 2N, how long will it take to come to rest? Given the resistance we can calculate the deceleration: 2 = 5a a = -0.4m/s2 Then use the knowns to solve: u = 2.64m/s v = 0 t = 6.6s Fnormal mg sin(a) mg cos(a) mg ao

Forces being pulled or pushed This guy is pulling a sled of salamis or fish or something across the ice at a constant speed. Which force, or forces, is he working against? Since he is pulling at an angle, he is working against the friction in the X direction and the weight in the Y direction.

Forces being pulled or pushed This guy is pulling a sled of salamis or fish or something across the ice at a constant speed. If the sled is 90kg, the angle 30o, and the friction provides resistance of 45N, then what would be the tension in the rope if the mass of the rope itself is ignored? Since friction only acts in the X direction we can find tension by 45 = Tcos30 T = 52N

Forces being pulled or pushed This guy is pulling a sled of salamis or fish or something across the ice at a constant speed. If the sled is 90kg, the angle 30o, and the friction provides resistance of 45N, then what would be the normal force? The normal force would no longer be equal to the weight since there is a y-component. The normal can be found by: mg – Tsin(a) = N 900 – 52sin(30) = 874N

Forces being pulled or pushed This guy is pulling a sled of salamis or fish or something across the ice at a constant speed. If the ice becomes more smooth and the friction drops to 20N, but the man continues to pull with the same force, what will happen? The man will accelerate at an acceleration given by: The resultant forward force would be 45 – 20 = 25N 25N = 90a a = 0.278 m/s2

Forces with pulleys In this situation the inclined plane is smooth, and the mass of P is great enough so that it will move down the ramp. What equation could be set to describe this system with relation to tension T, mass of p mp, gravity g and acceleration a? Given that the x-component moves parallel to the plane, the force mpg*sin(a) which moves down the gradient is moving against the tension T of the string. We also know that sin(a) = 0.7/2.5. Therefore: (0.7/2.5) mpg – T = mpa

Forces with pulleys In this situation the inclined plane is smooth, and the mass of P is great enough so that it will move down the ramp. Given that (0.7/2.5) mpg – T = mpa, what is the tension and the acceleration down the ramp if the mass of P and Q are 3kg and 0.6kg respectively? (0.7/2.5) * 3g – T = 3a T – 0.6g = 0.6a Solving for a 2.4 = 3.6a a = (2/3)m/s2 T = 6.4N

Forces with pulleys In this situation the inclined plane is smooth, and the mass of P is great enough so that it will move down the ramp. When Q hits the pulley the string breaks and P moves down the plane, reaching the bottom at a speed of 2m/s. How long is the string? First you should find the initial speed, which will be the speed of Q when it reaches the pulley. s=0.7, a=(2/3), u = 0, v = ? v2 = 2(2/3)(0.7) gives 0.967m/s Now, acceleration increases to g*sin(a) which is (0.7/2.5)g or 2.8m/s2 a = 2.8, u = 0.967, v = 2, s = ? 4 = 0.935 + 2 * (2.8) * s s = 0.55m which means the string is 1.95m