Newton’s Laws and Forces of Friction Problems and Solutions 3

 Problem#1

Materials such as automobile tire rubber and shoe soles are tested for coefficients of static friction with an apparatus called a James tester. The pair of surfaces for which µ_s is to be measured are labeled B and C in Figure 1. Sample C is attached to a foot D at the lower end of a pivoting arm E, which makes angle θ with the vertical. The upper end of the arm is hinged at F to a vertical rod G, which slides freely in a guide H fixed to the frame of the apparatus and supports a load I of mass 36.4 kg. The hinge pin at F is also the axle of a wheel that can roll vertically on the frame. All of the moving parts have masses negligible in comparison to the 36.4-kg load. The pivots are nearly frictionless. The test surface B is attached to a rolling platform A. The operator slowly moves the platform to the left in the picture until the sample C suddenly slips over surface B. At the critical point where sliding motion is ready to begin, the operator notes the angle θ_s of the pivoting arm. (a) Make a free-body diagram of the pin at F. It is in equilibrium under three forces. These forces are the gravitational force on the load I, a horizontal normal force exerted by the frame, and a force of compression directed upward along the arm E. (b) Draw a free-body diagram of the foot D and sample C, considered as one system. (c) Determine the normal force that the test surface B exerts on the sample for any angle θ. (d) Show that µ_s = tan θ_s. (e) The protractor on the tester can record angles as large as 50.2°. What is the greatest coefficient of friction it can measure?

Fig.1

Answer:
(a) See Figure (a) to the right.
(b) See Figure (b) to the right.
(c) For the pin,

∑F_y = ma_y

C cos θ – mg = 0

C = (36.4 kg)(9.80 m/s²)/cos θ = 357 N/cos θ

For the foot,

∑F_y = ma_y

+n_B – C cos θ = 0

n_B = C cos θ = 357 N

(d)  For the foot with motion impending,

∑F_x = ma_x

f_s – C sin θ = 0

µ_sn = C sin θ_s

µ_s  = C sin θ_s/C cos θ_s

µ_s = tan θ_s

(e) The maximum coefficient is

µ_s = tan 50.2⁰ = 1.20

Problem#2
What horizontal force must be applied to the cart shown in Figure 3 in order that the blocks remain stationary relative to the cart? Assume all surfaces, wheels, and pulley are frictionless. (Hint: Note that the force exerted by the string accelerates m₁.)

Fig.3

Answer:
∑F = ma

For m₁: T = m₁a
For m₂: T – m₂g = 0

Eliminating T,

a = m₂g/m₁

For all 3 blocks:

F = M_totala

F = (m₁ + m₂ M)m₂g/m₁

Problem#3
A student is asked to measure the acceleration of a cart on a “frictionless” inclined plane as in Figure 5, using an air track, a stopwatch, and a meter stick. The height of the incline is measured to be 1.774 cm, and the total length of the incline is measured to be d = 127.1 cm. Hence, the angle of inclination ! is determined from the relation sin θ = 1.774/127.1. The cart is released from rest at the top of the incline, and its position x along the incline is measured as a function of time, where x = 0 refers to the initial position of the cart. For x values of 10.0 cm, 20.0 cm, 35.0 cm, 50.0 cm, 75.0 cm, and 100 cm, the measured times at which these positions are reached (averaged over five runs) are 1.02 s, 1.53 s, 2.01 s, 2.64 s, 3.30 s, and 3.75 s, respectively. Construct a graph of x versus t2, and perform a linear least-squares fit to the data. Determine the acceleration of the cart from the slope of this graph, and compare it with the value you would get using a’ = g sin θ, where g = 9.80 m/s².

Fig.5

Answer:
Consider the table below

t(s)	t2(s2)	x(m)
0	0	0
1.02	1.040	0.100
1.53	2.341	0.200
2.01	4.040	0.350
2.64	6.970	0.500
3.30	10.89	0.750
3.75	14.06	1.00

From ½ at² the slope of a graph of x versus t² is ½ a, and

a = 2 x slope = 2(0.0714 m/s²) = 0.143 m/s²

from a’ = g sin θ

a’ = 9.80 m/s² (1.774/127.1) = 0.137 m/s², different by 4%

The difference is accounted for by the uncertainty in the data, which we may estimate from the third
point as

[0.350 – (0.0714)(4.04)]/0.350 = 18%

Problem#4
Initially the system of objects shown in Figure 6 is held motionless. All surfaces, pulley, and wheels are frictionless. Let the force F be zero and assume that m₂ can move only vertically. At the instant after the system of objects is released, find (a) the tension T in the string, (b) the acceleration of m2, (c) the acceleration of M, and (d) the acceleration of m₁. (Note : The pulley accelerates along with the cart.)

Answer:

For m₁ : T = m₁(a – a’)

a = T/m₁ + a’                    (1)

For M;  n = Ma’ = T          (2)

For m₂: m₂g – T = m₂a    (3)

(a) Substitute the value for a and a’ from (1) and (2) into (3) and solve for T:

m₂g – T = m₂[T/m₁ + T/M]
m₂g = T{1 + m₂[1/m₁ + 1/M]}

m₁m₂Mg = T{m₁M + m₂(m₁ + M)}

T = m₁m₂Mg/[m₁M + m₂(m₁ + M)]

(b) Solve (3) for a and substitute value of T:

m₂g – T = m₂a

m₂g – {m₁m₂Mg/[m₁M + m₂(m₁ + M)]}= m₂a

a = g – {m₁Mg/[m₁M + m₂(m₁ + M)]}

a = m₂g(m₁ + M)/[m₁M + m₂(m₁ + M)]             

(c) From (2), a’ = T/M, Substitute the value of T:

a' = m₁m₂g/[m₁M + m₂(m₁ + M)]

(d) then

a – a’ = m₂g(m₁ + M)/[m₁M + m₂(m₁ + M)] - m₁m₂g/[m₁M + m₂(m₁ + M)]

a – a’ = Mm₂g/[m₁M + m₂(m₁ + M)]   

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