Calculating Normal Force & its Relation to Incline Planes - Prof. Brian N. Espino, Study notes of Physics

Download Calculating Normal Force & its Relation to Incline Planes - Prof. Brian N. Espino and more Study notes Physics in PDF only on Docsity! Day 8, Ch 4 Forces Normal Forces Looking at Newton’s 2nd Law. F = ma A 10 kg book is sitting on a 20 kg desk. What normal force does the ground exert on the desk. Free body diagrams D B wb Fnb Fnd wd wb FP, is the opposite (parallel) component of the weight vector. It is parallel to the inclined plane. This is the component of the weight that causes the object to slide down the plane. FP = W sin = mg sin FN w FN FP FN FP w What happens when we increase from 0 to 90 degrees? Let mass = 50 kg. FP = (50 kg)(9.8m/s 2) sin FP = (50 kg)(9.8m/s 2) sin FP = (50 kg)(9.8m/s 2) sin FP = (50 kg)(9.8m/s 2) sin FP = (50 kg)(9.8m/s 2) sin FP = (50 kg)(9.8m/s 2) sin FN w FN FP Normal Forces (from Tuesday) What happens when we increase from 0 to 90 degrees? FN = (50 kg)(9.8m/s 2) cos FN = (50 kg)(9.8m/s 2) cos 0 = 490 N FN = (50 kg)(9.8m/s 2) cos 10 = 483 N FN = (50 kg)(9.8m/s 2) cos 30 = 424 N FN = (50 kg)(9.8m/s 2) cos 45 = 346 N FN = (50 kg)(9.8m/s 2) cos 60 = 245 N FN = (50 kg)(9.8m/s 2) cos 90 = 0 N Fn Free body diagram FN w Fnet Fnet is the total, or resultant force. It is directed along the plane. Applying the 2nd Law Fnet = ma = (50kg)a a = Fnet/m a = (50kg)g (sin 30)/(50 kg) a = 4.9 m/s2 Fnet Another way of looking at incline plane problems (rotating the coordinate system) Rotate coordinates by angle of incline, . Now the normal force is along the y-direction. You can see the x and y (parallel and normal components of the weight vector. FN w FN FP x y Free Body diagram The normal component of the weight is again balanced out by the normal force. Fnet is the parallel component of the weight. y x FN FP FN w FN w Vector addition Fnet First we use the force work to find: Fnet = ma a = a = g sin a = g sin 20 = 3.4m/s2 Now use the acceleration and solve for the velocity of the block after sliding 2 meters. = (0 m/s)2+ 2(3.4m/s2)(2m) Vf = 3.7 m/s (direction is down the incline) xavv f 2 2 0 2 m mg m Fnet )(sin What force (PULL)is needed to pull a sled up a frictionless hill at constant velocity? • mass of sled = 10 kg • Angle of hill is 45 degrees. 450 What force (PULL)is needed to pull a sled up a frictionless hill at constant velocity? • mass of sled = 10 kg • Angle of hill is 45 degrees. Since the velocity is constant, the sum of all the forces needs to be zero. So the PULL force needs to balance out the parallel component of the weight vector. PULL = mg (sin 45) = (10kg)g (sin 45) = 69 N 450 Free body diagram of sled PULLFN w 450 Friction How to calculate frictional force. Frictional force, Ff , depends on the material of the two surfaces involves. Given by coefficient of friction ( ). See table on page 101 for examples. There are two coefficients: static ( s) and kinetic ( k) Ff also depends on the normal force Static friction force Ff sFN Kinetic friction force Ff = kFN Static friction • Ff sFN • The static friction force can vary from 0 to sFN. • sFN is known as the maximum static friction force. • This is the force needed to be overcome to start sliding an object across the surface. • See figure 4.19 on page 101. Example What coefficient of friction is needed to keep a block from sliding down an incline? The static friction force balances out the parallel component of The weight. FN w FN FP FN and Fpush counteract each other in the horizontal direction. Weight and friction force balance each other out in the vertical direction. The friction is what is holding up the book. Let the book have mass 5 kg and s = .4. W = (5kg)g = 49 N For W = Ff, Ff = sFN = 0.4FN = 0.4 Fpush 49 N = 0.4 Fpush Fpush= 122.5 N FNFPush W Ff Kinetic Friction Problem You are pushing a block up an incline. The angle of inclination is 30 degrees. The mass of the block is 20 kg. The coefficients of friction are s = 0.3 and k = 0.2. What force must be applied to the box to keep the speed constant? Since block is sliding across surface use k = 0.2. Free body diagram Fapplied FN W Ff Since velocity is constant, a = 0. Fnet = ma = 0 We are only concerned with motion along the incline. Fapplied, Ff, and W have components along the incline. Writing 2nd Law along incline we get: Fapplied - Ff - mg sin 30 = ma = 0 Fapplied = Ff + mg sin 30 = kmg(cos 30)+mg(sin 30) Fapplied = 0.2(20kg)g(cos 30) + (20kg)g(sin 30) = 131 N Free body diagram FN W Ff Fapplied Incline + Jerking the scale upwards: The force exerted by the scale (T) on the mass is in the positive y-direction. W is in the negative y-direction. Acceleration is in the positive y-direction. 2nd Law: F = ma T – W = ma T = W + ma T= mg + ma = m(g+a) T W y Jerking the scale downwards: The force exerted by the scale (T) on the mass is in the positive y-direction. W is in the negative y-direction. Acceleration is in the negative y-direction. 2nd Law: F = ma T – W = -ma T = W – ma T= mg - ma = m(g – a) T W y Example: Mass 1 is on frictionless surface. Mass 2 is allowed to fall. What accelerations do the objects have. (They are tied to together.) m1 = 20 kg m2 = 10 kg Free Body Diagrams Mass 1 Mass 2 T Fn T W2W1