2. Box slides along horizontal at velocity constant. FfFf FpFp FwFw v c therefore,  F = 0 ; F p + F f = 0 ; F F = -F p Rest, therefore,  F = 0 ; F W + F N = 0 ; F N = -F w FNFN on parallel on perpendicular

3. Box slides along ho rizontal at velocity constant. FfFf FpFp FwFw on parallel: v c therefore,  F = 0 ; F p + F F = 0 ; F f = -F p F F = -(34N) = -34N What is  if the box is 12 kg and F p is 34 N? F w = mg = 12 kg x -9.8 m/s 2 = -120 N,  = F F /F N = -34N/120N  = 0.28 FNFN on perpendicular: rest, therefore,  F = 0 ; F w + F N = 0 ; F N = -F w F N = -(-120N) = 120N

4. Box slides along horizontal at velocity constant. FfFf FwFw v c therefore,  F = 0 ; F p + F F = 0 ; F F = -F p FaFa FpFp FvFv  FNFN on parallel: on perpendicular: rest, therefore,  F = 0 ; F W + F N + F v = 0 F N = -F w - F v This time the force (F a ) is applied at an upward angle on the box. Part of F a acts parallel to the ramp and part acts perpendicular. The parallel part of F a pulls the box forward and the perpendicular part pull the box upward.

5. Box slides along horizontal at velocity constant. FfFf FpFp FwFw On parallel: v c,  F = 0 F p + F F = 0 F F = -F p FaFa FvFv  What is  if a 65N force applied at a 30 0 slides a 12 kg box at a constant velocity? F v = Sin 30 0 (65N) = +32N F p = Cos 30 0 (65N) = +56N  = F F / F N = F F / (-F w – F v ) = -56N / -(-120N) – (32N)  FNFN on perpendicular: rest,  F = 0 F w + F v + F N = 0 F N = -F w – F v

6. Box accelerates along horizontal FfFf FpFp FwFw FNFN On parallel: accel,  F = ma What forces act in the direction of the acceleration? F p + F F = ma F p = ma - F F on perpendicular: rest,  F = 0 F w + F N = 0 F N = – F w.

7. Box accelerates along horizontal FfFf FpFp FwFw F p = -F f + ma F N = -F w F f =  F N What force is needed to accelerate the 12 kg box at 2.3 m/s 2 if “  ”  is 0.29? F N = -F w = -mg = -(12 kg x -9.8 m/s 2 ) = +120 N F p = 62N FNFN F p = -F f + ma = -  F N + ma = - -0.29(120N) + 12 kg(2.3 m/s 2 ) F f is negative because its left, opposite F p. F N isn’t negative But it has to be added to make F f negative!

8. Box accelerates along horizontal. FfFf FpFp FwFw F p + F f = ma F N + F w + F v = 0 FaFa FyFy What acceleration does a 65N force, applied at a 30 0 from the horizontal, give a 12 kg box if  is 0.29? F y = Sin 30 0 (65N) = +32N F x = Cos 30 0 (65N) = +56N F p + F f = ma Solve for “a” a = F p + F F / m F N = -(-120N) - (32N) = 88N a = F p +  F N / m a = F p +  F w -F v ) / m a = 56N + -(0.29)(88N) / 12kg = 2.5 m/s 2 F N = -F w - F v

9. Box slides down the ramp at a constant velocity FwFw FfFf FpFp  F F = -F p F N ’ = Cos  F w FpFp FN’FN’  F p = Sin  F w FNFN F N = -F N ’ on parallel on perpendicular V c  F = 0 F F + F p = 0 Rest;  F = 0 F N + F N ’ = 0

10. Box slides down the ramp at a constant velocity F f = -F p F N = Cos  F w FwFw FfFf FpFp  FpFp FN’FN’  F p = Sin  F w What is  if a 12 kg box slides down a 6.5 m ramp that is 3.9 m high at a constant velocity? Sin  opp/hyp = 3.9 m/6.5 m    = 0.75 F N = -F N ’  F f /F N = -F p /F N = - Sin 35 0 (12kg)-9.8m/s 2 /Cos 37 0 (12kg)-9.8m/s 2

11. Box accelerates down the ramp F p + F f = ma F N ’ = Cos  F w FwFw FfFf FpFp  FpFp FN’FN’  F p = Sin  F w What acceleration does a 12 kg box have if  is 0.45 and the ramp is 6.5 m long and 3.9 m high? Sin  opp/hyp = 3.9 m/6.5 m   F p + F f = ma a = F p + F f /m F p = Sin 37 0 (12kg)-9.8m/s 2 = -71N F N ’ = Cos 37 0 (12kg)-9.8m/s 2 = -94N F f =  F N F f = 0.45(94N) = 42N a = -71N + (42N)/ 12kg = -2.4 m/s 2 on parallel Accel.;  F = ma

12. Box accelerates up the ramp F a + F p + F f = ma F N ’ = Cos  F w FwFw FfFf FpFp  FpFp FN’FN’  F p = Sin  F w What acceleration does a 12 kg box have up the ramp if a force of 145 N is applied up the ramp and  is 0.45 and the ramp is 6.5 m long and 3.9 m high? Sin  opp/hyp = 3.9 m/6.5 m   F a +F p + F f = ma a = F a + F p + F f /m F p = Sin 37 0 (12kg)-9.8m/s 2 = -71N F N ’ = Cos 37 0 (12kg)-9.8m/s 2 = -94N F f =  F N F f = 0.45(94N) = -42N a = 145 N +(-71N) + (-42N)/ 12kg on parallel Accel.;  F = ma FaFa a = 2.7 m/s 2

13. FfFf FwFw FNFN Friction Stopping an Object A 55 kg person running at 6.6 m/s trips and falls and skids to rest in 2.1 m. What is  ? Note- there is no forward force on the person once he trips and starts sliding!!! on perpendicular on parallel Rest, therefore,  F = 0 ; F W + F N = 0 F N = -F w F N = -(-540 N) = 540 N accel,  F = ma F F = ma  F N = ma  = ma/F N = 55 kg (10 m/s 2 )/540 m  = 1.0 -540 N The person decelerates once he Hits the ground because of Friction. We know v 1 = 6.6 m/s, v 2 = 0, and d = 2.1 m, so “vad” will give us the acceleration. a = v 2 2 – v 1 2 /2d a = (6.6 m/s) 2 – 0/2(2.1m ) = 10 m/s 2

14. Motion is in the vertical, so the parallel plane is also in the vertical. Lifting an object An 11.0 kg box is accelerated upward at 4.40 m/s 2. What force is needed to do this? F w = -108 N F a = ? Note, there are no forces On the horizontal!!! on parallel accel,  F = ma F w + F a = ma F a = ma – F w F a = (11.0 kg x 4.40 m/s 2 ) – (-108 N) = 156 N