Take a look at this component-of-force thing. When a block sits on an incline, we can look at the forces on it in two ways:

or

To calculate the magnitude of the two components, study this:
Having trouble swallowing the equality of the two angles? Check out these diagrams.
So now look at what this says about the magnitude of the forces:
Since the incline presses perpendicularly against the block just enough to cancel out the WcosØ, the only remaining force acting is WsinØ.

29. A 6 pound force pushes a 20 pound weight down a 19° frictionless slope. What is its acceleration?

F = ma 6 + 20sin(19) = (20/32)a a = 20.0 ft/sec2

30. Find the unknown in each case.

a)	b)
a) F = ma 11 * 6 - 80sin(27) = (80/9.8)a a ~= 3.64 m/sec2	b) F = ma 30 + m * 9.8sin(22) = 7m 30 = m[7 - 9.8sin(22)] m ~= 9.01 kg
c)	d)
c) F = ma k * 9 - 3 * 32sin(57) = 3 * 5 a ~= 10.6 ft/sec2	d) F = ma 17 + 9 * 9.8sin(18) - 6 = 9a a ~= 4.25 m/sec2

31. A 23 kg mass is pulled by a spring (spring constant 9 nt/cm) stretched 13 cm and pushed by a 43 nt force pointed at a 40° angle. If the incline is banked at a 28° angle, what is the acceleration?

F = ma 43cos(40) + (9 * 13) - 23 * 9.8sin(28) = 23a a ~= 1.92 m/sec2

32. And this...just write a true F = ma equation, but don't solve it for anything.

F = ma Fcos(Ø1) + mgsin(Ø2) + kx = ma

We have, up to now, dealt only slightly with the interaction of one mass with another, yet forces between masses are of fundamental importance in figuring out the real world. Try some more.

33. Daring space traveler Andy Jestion notes 48 and 35 kg masses being hauled behind a small space vehicle at 3.6 m/sec².

a) Draw the forces acting on each mass. b) Write the F = ma equation for each. c) Solve for the two tensions in the tethers.

a) Force diagrams:

b) F = ma T1 = 48(3.6) T2 - T1 = (35)(3.6)

c) Add the equations to cancel T1: T1 = 48(3.6) T2 - T1 = (35)(3.6) Solve for T2: T2 = 299 nt Plug T2 into the second equation and solve for T1: 299 - T1 = (35)(3.6) T1 = 173 nt

In this last problem, I introduced variables T₁ and T₂. In doing this I invoked the fundamental principle of algebra, which is rarely explained to students in their algebra classes. So, lest you wait any longer, HERE IT IS:

Thou Shall Replace Ignorance with a Letter and Proceed.

It is remarkable how powerful this idea is. Rather than stopping to puzzle over an unknown, just slap a symbolic name on it and continue to write true things.

Try that idea on a few more situations.

34. Three masses are tethered together as shown. The tension in the lead string is 44nt.

a) What is the acceleration of the system? b) What are the other tensions?

a) Create equations for each mass: F = ma 44 - T2 = 7a T2 - T1 = 4a T1 = 6a Add the equations together (T1 and T2 cancel): 44 = (7 + 4 + 6)a a ~= 2.59 m/sec2

b) Plug in the acceleration to find the tensions: 44 - T2 = 7(2.59) T2 ~= 25.9 nt T1 = 6(2.59) T1 ~= 15.5 nt

35. Two masses are separated by a spring of 244 nt/m stiffness, and a 90 nt force is applied as shown. Again, draw the forces on each mass and determine how far will the spring compress.

Create the two equations: F = ma 90 - 244x = 12a 244x = 7a Add the equations together: 90 = 12a + 7a a ~= 4.74 m/sec2 so, 244x = 7(4.74) x ~= .136 m

36. 160 lb Malcolm Tent is shoved forward by a force of 75 lb. He is holding a 13 lb birthday present. What horizontal force acts between Malcolm and the present? (One might also ask what force acts between Malcolm and the past.)

Here's one approach: F = ma 75 = [(160 + 13)/32]a a ~= 13.9 ft/sec2 So, it could be solved like: F = ma 75 - F = (160/32)(13.9) F ~= 5.64 lb

37. Given the masses and accelerations, find the tensions.

a)	b)	c)	d)
a) F = ma T - 11 * 9.8 = 11 * 3 T = 141 nt	b) F = ma but mg = (4/3)(3.14159)(6/12)3(6)(32) mg = 100.5 lb 100.5 - T = (3.14 slug)(7) T ~= 78.5 lb	c) F = ma T - (85 * 980) = 85 * 110 T ~= 9.27 * 104 dyne	d) F = ma 9.8 * 8 - T = 8 * 9.8 T = 0 nt

38. A 73 kg man wishes to lower himself from a 36 m high apartment that is burning out of control. He has a rope which will support only 700 nt.

a) At what acceleration can he lower himself without the rope breaking? b) With what speed will he hit the ground?
a) F = ma a = F/m [(73 * 9.8) - 700]/73 a ~= .211 m/sec2 b) v2 = 2as 2(.211)(36) = 15.2 v ~= 3.90 m/sec

39. A 45 kg crate of pickled herrings is to be rapidly lifted to 13.0 m from the ground. What tension should be applied to the rope to lift the crate in 4.0 sec?

First find the acceleration: S = (1/2)at2 a = 2s/t2 a = (2 * 13)/42 a ~= 1.625 m/sec2 But F=ma T - (45 * 9.8) = (45)(1.625) T ~= 514 nt

40. A macho-crazed mountaineer slides down a rope. If his weight is 815 nt, and he applies a constant force of 55 nt to the rope, what is his downward acceleration?

F = ma a = F/m a = (815 - 55)nt/(815/9.8)kg a ~= 9.14 m/sec2 Note that the resistance to acceleration comes from an objects mass, not its weight. Remember that F = ma

41. Two small rocket engines of 8 nt and 14 nt thrust are attached to a 2 kg organically grown grapefruit as shown. With what acceleration, and in what direction will the grapefruit move?

Break the problem into vertical and horizontal components: F = ma a = F/m V) ay = Fy/m ay = 8 nt/2 kg ay = 4 m/sec2 H) ax = Fx/m ax = 14 nt/2 kg ax = 7 m/sec2 Using trig, we can find the the vector sum of the two components (hypotenuse) and the angle. a2 = b2 + c2 a ~= 8.06 m/sec2 tan(Ø) = 4/7 Ø = 29.7°

42. Now find the accelerations and directions of these if they are in weightless space.

a)	b)	c)
a) F = ma a = (2002 + 3002)1/2/80 a = 4.51 cm/sec2 tan(Ø) = 200/800 Ø = 33.7°	b) F = ma a = (302 + 372)1/2/1.4 a = 34.0 ft/sec2 tan(Ø) = 37/30 Ø = 51.0°	c) F = ma a = (122 + 212)1/2/[(4/3)(3.14159)(.14)3(348)] a = 6.05 m/sec2 tan(Ø) = 21/12 Ø = 60.3°

43*. Think back to how we solved statics problems when you do these. Ignore gravity.

a)	b)
a) H) Fx = 6cos(50) Fx = -4.14 nt V) Fy = 6sin(50) Fy = 4.60 nt F = ma a = (-4.142 + 4.602)1/2/7 a = .884 m/sec2 tan(Ø) = 4.6/4.14 Ø = 48.0°	b) H) Fx = 20cos(23) - 8cos(50) Fx = 13.27 lb V) Fy = 20sin(23) + 8sin(50) Fy = 13.9 lb F = ma a = (13.272 + 13.92)1/2/2 a = .9.61 ft/sec2 tan(Ø) = 13.9/13.27 Ø = 46.3°

Keep in mind, when applying Newton's second law, that the F is the net, or unbalanced force acting on the mass.

44. A 65 kg man on a skateboard slows from 2.0 m/sec to 0.8 m/sec in 4.5 sec. What is his acceleration? What frictional force is slowing him?

a) a = Ç(v)/Ç(t) a = (.8 - 2.0)/4.5 a ~= -.267 m/sec2

b) F = ma F = (65 kg)(-.267 m/sec2) F ~= -17.3 nt

45*. A car starts from rest, 50 ft from a cliff. It accelerates at 19 ft/sec² until it goes over the cliff to the valley 85 ft below. How far from the base does it land?

a) First find the speed at the edge: v2 = 2as v2 = 2(19 ft/sec2)(50 ft) v ~= 43.6 ft/sec Next, find the time in the air: s = (1/2)at2 t = (2s/g)1/2 t = [(2 * 85)/32]1/2 t ~= 2.30 sec Finally, find the distance from the base: s = vt s = (43.6)(2.3) s ~= 100 ft

46*. A stunt man and his moped mass a combined 408 kg. If he plans to jump the 17 m gap to the 4 m lower cliff on the opposite side,
a) How fast must he be going when he takes off?
b) How fast must his motorcycle accelerate?
c) What force must the wheels supply?

a) V) s = (1/2)at2 t = (2s/g)1/2 t = [(2 * 4)/9.8]1/2 t = .904 sec H) s = vt v = s/t v = 17 m/.904 sec v ~= 18.8 m/sec

b) v2 = 2as a = v2/2s a = 18.82/(2 * 12) a ~= 14.7 m/sec2

c) F = ma F = (408 kg)(14.7 m/sec2) F = 6.00 * 103 nt