Pulleys enter dynamics under the same rules we had in statics; they are massless and frictionless. This may sound phony, but it's not a bad approximation of reality. We continue, then, to conclude that tensions on both sides of a pulley are equal, and the single rope has twice the tension as the double rope.

We now must note that the double rope moves (and therefore accelerates) twice as far and twice as fast as the single rope:

33. Rex Things accelerates his car at 7 ft/sec², pulling a cart behind him with the pulley system shown. What is the tension in the rope?

First recognize that the cart must accelerate at 3.5 m/sec2 and the rope attached to it has a tension of 2T F = ma 2T = 13 * 3.5 T ~= 22.3 lb

34. Find the acceleration of m₁.

First create the force equations for each mass: equation #1 F = ma T/2 - µ2m2g = m2(2a) equation #2 F = ma F - T - µ1m1g = m1a Double the first equation (so T's will cancel) and add to the second: T - 2µ2m2g = 4m2a F - T - µ1m1g = m1a ------------------------ F - µ1m1g - 2µ2m2g = 4m2a + m1a) a = (F - µ1m1g - 2µ2m2g)/(4m2 + m1)

35. Two masses (m₁ and m₂) are attached to two pulleys which are connected by a common string. If m₁ is connected to one end of the string while m₂ is connected to the center of the second pulley, what is the acceleration of m₁?

F = ma equation #1 T - m1g = m1a equation #2 m2g - 2T = m2(a/2) Double the first equation (so T's will cancel) and add to the second: 2T - 2m1g = 2m1a m2g - 2T = m2(a/2) ------------------ m2g - 2m1g = [2m1 + (m2/2)]a a = (m2 - 2m1)g/[2m1 + (m2/2)]

36*. There is no limit to the richness of these problems. Solve for a.

F = ma equation #1 T - m1g = m1a equation #2 T2 - T1 - µm2g = m2a equation #3 m3g - (T2/2) = m3(2a) Double the third equation (so T's will cancel) and add: -m1g - µm2g + 2m3g = (m1 + m2 + 4m3)a a = (-m1 - µm2 + 2m3)g/(m1 + m2 + 4m3)

37*. Find the acceleration of m₁.

F = ma equation #1 m1g - T1 = m1a equation #2 2T1 - T2 - µ2m2g = m2(a/2) equation #3 2T2 - m3gsin(Ø) - µ3m3gcos(Ø) = m3(a/4) Double the second equation, quadruple first equation (so T's will cancel), and add: 4m1g - 2µ2m2g - m3gsin(Ø) - µ3m3gcos(Ø) = [4m1 + m2 + (m3/4)]a a = [4m1 - 2µ2m2 - m3(sin(Ø) + µ3cos(Ø))]g/[4m1 + m2 + (m3/4)]

38. A 5 kg block moving at 6.6 m/sec² encounters a frictional (µ = .3). How far will it slide before coming to rest?

The block feels a retarding force equal to µmg. So it accelerates at: a = F/m a = µmg/m a = µg a = (.3)(9.8) a ~= 2.94 m/sec2 But: v2 = 2as + vo2 02 = 2(-2.94)s + 6.62 s ~= 7.41 m

39. A 140 kg refrigerator with coefficient of friction 0.53 is on the bed of a pickup. What is the maximum acceleration the pickup can have without the refrigerator sliding?

The refrigerator is accelerated forward by friction with the truck bed. The maximum force friction can apply is µmg. a = F/m a = µmg/m a = µg a = (.53)(9.8) a ~= 5.19 m/sec2

40. A 4.4 kg cart with frictionless wheels carries a 6.9 kg mass (µ = .2). What is the maximum force that can be applied to it without causing the block to slip?

The block is accelerated forward by friction from the cart. Thus the maximum possible acceleration is: a = F/m a = µmg/m a = µg But F must accelerate the block and the cart, so: F = ma F = (6.9 + 4.4)(.2 * 9.8) F ~= 22.2 nt If a greater force is applied, the cart will slip out from under the block.

Please note that a slipping block is just a complicated way of stating the acceleration of the system. From now on you should see a slipping block as an obscure way to tell you that the acceleration is just µg.

41. Barb Dwyer pushes on the 8.0 kg block with a force of 53 nt. What force acts between the blocks?

Define the force f between the blocks. equation for 6kg block f - µmg = ma f - .2(6 * 9.8) = 6a a = (1/6)[f - .2(6 * 9.8)] equation for 8kg block F - f - µmg = ma 53 - f - .3(8 * 9.8) = 8a Plug in the acceleration from the first equation(6kg) into the second equation(8kg): 53 - f - .3(8 * 9.8) = 8 * (1/6)[f - .2(6 * 9.8)] 53 - .3(8 * 9.8) + (8/6) * .2(6 * 9.8) = f + (8/6)f f ~= 19.4 nt

We come, now, to the famous milk shake problem, best described in the Saturday Night Parable.

Betty Wohnt is out on her first date with Scotty Beameyup. She has just placed her strawberry shake on the dashboard when Scotty's archrival, Frank Zenbeens pulls up to the light. The inevitable macho showdown ensues with Betty drenched in shake and Scotty, having never taken physics, committing the social gaffe of his high school career.

With a little reasoning you can spare yourself the pain of falling into social oblivion. First let's look at the milk shake from the sidewalk that is, from a stationary frame of reference. The dashboard exerts a forward force on the shake. Since it is the only force acting in the line of acceleration, its magnitude is ma (since F = ma).
As the block is accelerated it rocks back so that its vertical support is applied at its rear end. Thus there are two torques applied about the center of the block: t1 = mg(w/2) (clockwise torque) t2 = ma(h/2) (counter-clockwise)

As long as the acceleration is small, ma is small and the shake does not rotate counter-clockwise. When the two torques are equal, the shake is just ready to go over.

mg(w/2) = ma(h/2) or a/g = w/h.

These are the true forces on the block, but they are somewhat messy to deal with mathematically, and they are not the forces we feel when we are accelerated with the shake. You'll find it easier to imagine the (fictitious) forces that seem to act on the block:

stable	unstable	metastable

From the last diagram we see the condition for just barely tipping is that

ma ---- mg

=

w --- h

so

a ---- g

=

w --- h

42. A rectangular shake, 30 cm high and 15 cm wide, rests on a dashboard. At what acceleration will it tip? (If you feel uneasy about not knowing its mass, assign it a value of 0.4 kg. It will cancel out anyway.)

ma ---- mg = w --- h .4a ---- .4 * 9.8 = 15 --- 30 a = (15/30) * 9.8 a = 4.90 m/sec2

43. A 2.8 kg block topples over at 4.1 m/sec². If the block is 8 cm wide, how tall is it?

ma ---- mg = w --- h h = g(w/a) h = 9.8(8/4.1) h ~= 19.1 cm

44. What is the maximum force that can be applied to the cart without having the block tip over?

ma ---- mg = w --- h a = g(w/h) a = 32(2/3) F = ma F = [3 + (128/32)][32 * (2/3)] F ~= 149.3 lb

Note that again we have an obscure way of knowing the acceleration of a system. This time a/g = w/h.

45. Arthur Podd wishes to push the cart shown in his fraternity's fifth annual Trash Can Race. What is the minimum possible time for the can to make it the required 10 meter distance without tipping?

The masses are irrelevant to this problem, though they are relevant to how hard Arthur must push. As before, the acceleration is fixed by the dimensions of the trash can. a ---- g = w --- h a = g(w/h) a = 9.8(.55/1.2) a ~= 4.49 m/sec2 But s = (1/2)at2: t = (2a/a)1/2 t = [(2 * 10)/4.49]1/2 t ~= 2.11 sec

46**. It is found that a force F will just topple a cylinder positioned on a cart. All that is known about it is its radius r, and that it has half the mass of the cart. Find the density of the cylinder in terms of F, r, and g.

Mass of Cylinder = pV = p(3.1415)r2h where h is equal to the height of the object. Mass of cart = 2p(3.1415)r2h Write two equations for acceleration: equation #1 a = g(w/h) a = g(2r/h) equation #2 F = ma a = F/m a = F/[3p(3.1415)r2h] Set each equation equal to each other: F/[3p(3.1415)r2h] = g(2r/h) p ~= F/[6(3.1415)r3g]

47. Juan Formababee, a proprietor of a local public house, notices that when he scoots a mug down the counter, the contents make a 5.0° angle with the horizontal. What is the coefficient of friction between counter and mug?

Consider a water molecule at the surface. It is affected by two forces, mg and the buoyant force of the water under it. The result of these forces is a horizontal backwards force which slows the molecule down. Based on this knowledge, we can say: tan(Ø) = (ma)/(mg) tan(Ø) = a/g tan(Ø) = (µg)/g tan(Ø) = µ µ ~= .0875

48. Mass m₁ is placed on an inclined plane as shown. All surfaces are frictionless, but m₁ is kept in place by accelerating the block to the right. What force must be applied to make this work?

The resultant force on m, must be perpendicular to the plane, otherwise m will slip. tan(Ø) = (ma)/(mg) F = ma F = mgtan(Ø) F = (m1 + m2)gtan(Ø)

49. An inquisitive bus rider dangles a yo-yo as his bus roars out of its parking place. If the bus accelerates at 2.2 m/sec², what angle will the yo-yo make with the vertical?

The yo-yo will hang in what seems to it a vertical direction. Thus we can find the angle using values we already have: tan(Ø) = (ma)/(mg) tan(Ø) = a/g tan(Ø) = 2.2/9.8 Ø ~= 12.7°

50*. Mass m₁ slides to the left with an initial velocity v₀. How far will it get before stopping?

Write the force equations for the two masses: T = µm1g = m1a m2g - 2T = m2(a/2) Note that m2 is moving up, but since it is slowing, it is accelerating down. Next, Double the first equation and add: 2µm1g + m2g = 2m1a + m2(a/2) 2µm1g + m2g = a(2m1 + (m2/2) a = [(2µm1 + m2)g]/(2m1 + (m2/2) But v2 = 2as + v02 where v = 0 s = v02/(2a) s = (v02/2) * [[(2µm1 + m2)g]/(2m1 + (m2/2)]

51*. m₂ slides down the plane. Find its acceleration.

Write the force equations for the two masses: F = ma 2T - Ff - Fg = ma 2T - µ1m1gcos(Ø1) - m1gsin(Ø1) = m1(a/2) F = ma Fg - Ff - T = ma m2gsin(Ø2) - µ2m2gcos(Ø2) - T = m2a Double the second equation and add: 2m2gsin(Ø2) - 2µ2m2gcos(Ø2) - µ1m1gcos(Ø1) - m1gsin(Ø1) = m1(a/2) + m2a a = g[2m2[sin(Ø2) - µ2cos(Ø2)] - m1[sin(Ø1) + µ1cos(Ø1)]]/[(m1/2) + 2m2]

52**. What force is necessary to accelerate m₂ at rate a?

Carefully study the forces on each mass to create the two force diagrams: F = ma 2T - µ1m1g = m1(a/2) F = ma F - µ1m1g - µ2(m1 + m2)g - T = m2a Multiply the second equation by 2 and add: 2F - 2µ1m1g - 2µ2(m1 + m2)g - µ1m1g = m1(a/2) + m2a F = µ1[(3/2)m1]g + µ2(m1 + m2)g + [m2 + (m1/4)]a

53*. What mass, m₂, will be small enough to allow m₃ to just barely tip over?

The dimensions of the top block determine a. ma ---- mg = w --- l Formulate the two force equations: Find the acceleration needed to tip over m3: a = (w/l)/g F = ma T - Ff = ma T - µ(m2 + m3)g = (m2 + m3)[(w/l)g] F = ma Fg - T = ma m1g - T = m1[(w/l)g] Add the two equations: m1g - - µ(m2 + m3)g = (m1 + m2 + 3)[(w/l)g] m2 = [m1[l - (w/g)] - m3[µ + (w/l)]]/[µ + (w/l)]