Newton's second law
For each box we have F = ma. The magnitude of a is the same for each box.
Mg – T = Ma
T – f = ma
f = mmg
a = (M – mm)g/(M + m), T = Mg – Ma = Mmg(1 + m)/(m + M)
None
Problem 1:
A box of mass m slides across a horizontal table with coefficient of friction m. The box is connected by a rope which passes over a frictionless pulley to a body of mass M hanging along side the table. Find the acceleration of the system and the tension in the rope.
Solution:
| Concepts, principles, relations
that apply to the problem: Newton's second law |
| Why do they apply? For each box we have F = ma. The magnitude of a is the same for each box. |
| How do they apply? Mg – T = Ma T – f = ma f = mmg a = (M – mm)g/(M + m), T = Mg – Ma = Mmg(1 + m)/(m + M) |
| Details of the calculation: None |
Problem 2:
In the figure below the coefficient of friction is the same
at the top and the bottom of the 700-g block.
(a) Draw a free body diagrams for both the 200-g and the
700-g blocks, considering all forces.
(b) If the acceleration is a = 70 cm/s2 when F = 1.3 N, how large is
the coefficient of friction
Solution:
| Concepts, principles, relations
that apply to the problem: Newton's second law |
| Why do they apply? This is a simple Newtonian Mechanics problem. |
| How do they apply? (a)
(b) F – T – f1 – f2 =
Ma |
| Details of the calculation: None |
Problem 3:
(a) An elevator in which a woman is standing moves upward at 4 m/s. If the woman drops a coin from a height 1.4 m above the elevator floor, how long does it take the coin to strike the floor? What is the speed of the coin relative to the floor just before impact? (b) Now assume that the elevator is moving downward with zero initial velocity and acceleration of 1 m/s2 at t = 0, the women releases the coin at t = 1 sec. How long does it take the coin to strike the floor? What is the speed of the coin relative to the floor just before impact?
Solution:
| Concepts, principles, relations
that apply to the problem: The equations of motion in an inertial and in an accelerating frame. |
| Why do they apply? We are suppose to analyze a coin drop in an inertial and in an accelerating frame |
| How do they apply? (a) The elevator is an inertial frame. The time it takes the coin to reach the floor is t = (2h/g)1/2 = 0.53s. The speed just before impact is v = gt = 5.24 m/s. (b) Now the elevator is an accelerating frame. In non-inertial frames fictitious forces appear. Consider a particle moving with velocity v in a reference frame K which moves with velocity V(t) relative to the inertial frame K0. The equations of motion are mdv/dt = -¶U/¶r - mdV/dt . Here dv/dt = g - 1 m/s2 = 8.8 m/s2 = g’, downward. t = (2h/g’)1/2 = 0.56s. The speed just before impact is v = g’t = 4.96 m/s. |
| Details of the calculation: None |
Problem 4:
Two trucks are parked back to back in opposite directions on a straight, horizontal road. The trucks quickly accelerate simultaneously to 3.0 m/s in opposite directions and maintain these velocities. When the backs of the trucks are 20 meters apart, a boy in the back of one truck throws a stone at an angle of 40 degrees above the horizontal at the other truck. How fast must he throw, relative to the truck, if the stone is to just land in the back of the other truck?
Solution:
| Concepts, principles, relations
that apply to the problem: Projectile motion |
| Why do they apply? The stone's motion is motion in two dimensions with constant acceleration. |
| How do they apply? Put the origin of the coordinate system at the back of the parked trucks and let the boy be in the truck moving into the negative x direction. When each truck has moved 10 m, is speed is v = 3m/s. Let v be the speed of the stone relative to the truck. The x-component of the stone’s speed with respect to the ground is vx = vcos40o – 3 m/s. The y-component of the stone’s speed with respect to the ground is vy = vsin40o. For the stone is to just land in the back of the other truck we need vxt = 20 m + (3 m/s)t. 0 = vyt – ½gt2, t = 2vy/g. Therefore 2vxvy/g = 20 m + (3 m/s)2vy/g. vxvy = g10 m + (3 m/s)vy. (vcos40o – 3 m/s)vsin40o = g10 m + (3 m/s)vsin40o. v2cos40osin40o = g10 m + (6 m/s)vsin40o. v2 = 199 (m/s)2 + 7.83 (m/s)v. v = 18.56 m/s. |
| Details of the calculation: None |
Problem 5:
A rock is launched from the ground level at a
speed v directed at an angle q with the
horizontal. It is noticed that after some (unknown) time t
after the launch, the distance between the rock and the launch
point begins to decrease.
(a) Find the smallest launch angle q
consistent with this observation.
(b) Find t, neglecting the air resistance.
Solution:
| Concepts, principles, relations
that apply to the problem: Projectile motion |
| Why do they apply? The roc is a projectile. |
| How do they apply? For projectile motion we have x = v cosq t, y = v sinq t - (1/2)gt2. |
| (a) The distance between the
rock and the launch
point is D = (x2 + y2)1/2 = (v2t2
+ g2t4/4 - vgt3sinq)1/2. For 0 < q < 90o and v > 0 the distance D increases initially. To find extrema for D we set dD2/dt = t(g2t2 - 3vgt sinq + 2v2). = 0. t2 - 3(v/g) sinq t + 2v2/g = 0, t = [3v/(2g)][sinq ± (sin2q - 8/9)1/2] If sin2q ³ 8/9 or q > 70.5287o, then two extrema exist. The figure below shows a plot of D versus t for v = 10 m/s and a launch angle smaller, equal, or greater than the critical angle q > 70.5287o.  The smallest launch angle q consistent with the observation is just greater than 70.5287o. (b) t = t1 = [3v/(2g)][sinq - (sin2q - 8/9)1/2] is the maximum we are looking for. For t > t1 D decrease and for t > t2 = [3v/(2g)][sinq + (sin2q - 8/9)1/2] it decreases again. |
| Details of the calculation: None |
Problem 6:
Consider a pendulum consisting
of a rigid thin rod with length L and negligible mass supporting a ball of mass
M. The pendulum is immersed in a viscous medium which causes a frictional force
F whose magnitude is proportional to the speed v of the ball,
F = -mv. It swings in a vertical plane
under the influence of gravity.
(a) Derive the equation of
motion of the pendulum, allowing for arbitrary angles
q of deflection from the vertical axis.
(b) Determine the fixed
points for which d2q/dt2
= 0 when dq/dt = 0. Determine for each fixed point
the critical value of the drag coefficient m
above which there is no oscillation about the point for small displacements.
Solution:
| Concepts, principles, relations
that apply to the problem: Newtonian mechanics |
| Why do they apply? Newton's second law gives the equation of motion. We are asked to find the equilibrium points, where the acceleration is zero. The equation of motion will tell us if small displacements away from an equilibrium point will lead to oscillations. |
| How do they apply? F = Ma (a) Equation of motion: MLd2q/dt2 = -Mgsinq – mLdq/dt d2q/dt2 + (m/M)dq/dt + (g/L)sinq = 0 (b) Fixed points: q = 0 and q = p. For small deviations from q = 0 we have d2q/dt2 + (m/M)dq/dt + (g/L)q = 0. q = q0exp(iwt), w2 - iwm/M - g/L = 0, w = im/(2M) ± (-m2/(4M2) +g/L)1/2 = ia ± b. (The displacement as a function of time is given by the real part of this solution.) If m < 2M(g/L)1/2, then b is real and q = q0exp(-at) exp(±ibt), we have oscillations. If m > 2M(g/L)1/2, then b is imaginary and we have no oscillations. For small deviations from q = p we have d2f/dt2 + (m/M)df/dt - (g/L)f = 0, where f = p + q. f = f0exp(lt), l2 + lm/M - g/L = 0, l = -m/(2M) ± (m2/(4M2) + g/L)1/2. l is always real, we have no oscillations for any value of m. |
| Details of the calculation: None |
Problem 7:
A child slides down a frictionless slide as shown in
figure.
(a) What is the minimum value of R for the child to not
immediately loose contact with the section of Radius R?
(b) If R is larger than that minimum value at what height h
will the child loose contact with the section of radius R?
(c) If R is smaller than that minimum value, how far from the end of the section with radius R will the child land?
Solution:
| Concepts, principles, relations
that apply to the problem: Energy conservation, centripetal acceleration |
| Why do they apply? Energy conservation yields the speed of the child when it reaches the top of the section with radius R. The centripetal acceleration needed to move in a circular path of radius R is v2/R. If g > v2/R then gravity can provide the centripetal acceleration. |
| How do they apply? (a) When the child reaches the top of the section with radius R it has kinetic energy ½mv2 = mg(H-R), and it is moving in the positive x-direction. The centripetal acceleration needed to move in a circular path of radius R is v2/R. If g > v2/R then gravity can provide the centripetal acceleration. We need 1 > 2(H-R)/R, R > (2/3)H for the child to stay in contact with the section with radius R. (b) The child’s velocity as a function of h is given by v = (2g(H – h))1/2. As long as gh/R > 2g(H-h)/R gravity can provide the centripetal acceleration needed for the child to move in a circular path. When h = 2H/3, the child will loose contact with the circular section. (c) If R < (2/3)H: x + R = (2g(H-R))1/2t, t = (2R/g)1/2, x = 2((R(H - R))1/2 – R. The child will land a distance x = 2((R(H - R))1/2 – R from the end of the section with radius R. |
| Details of the calculation: None |
