Body A in Fig. weighs $102$ N, and body B weighs $32$ N. The coefficients of friction between A and the incline is $μ_{s} = 0.56$ and $μ_{k} = 0.25$ . Angle $θ$ is $40^{\circ}$ . Let the positive direction of an x-axis be up the incline. In unit-vector notation, what is the acceleration of A if A is initially (a) at rest, (b) moving up the incline, and (c) moving down the incline?

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Solution

First, we check to see whether the bodies start to move. We assume they remain at
rest and compute the force of (static) friction that holds them there, and compare its
magnitude with the maximum value $μ_{s} F_{N}$ . The free-body diagrams are shown below.
T is the magnitude of the tension force of the string, f is the magnitude of the force of
friction on body A, $F_{N}$ is the magnitude of the normal force of the plane on body A, $m_{A} \to g$ is the force of gravity on body A (with magnitude $W_{A} = 102 N$ ), and $m_{B} \to g$ is the force of
gravity on body B (with magnitude $W_{B} = 32$ N). $θ = 40^{\circ}$ is the angle of incline. We are
not told the direction of $\to f$ but we assume it is downhill. If we obtain a negative result for
f, then we know the force is actually up the plane.
(a) For A we take the +x to be uphill and +y to be in the direction of the normal force. The
x and y components of Newton’s second law become
$T - f - W_{A} sin θ = 0$
$F_{N} - W_{A} cos θ = 0$
Taking the positive direction to be downward for body B, Newton’s second law leads to $W_{B} - T = 0$ . Solving these three equations leads to
$f = W_{B} - W_{A} sin θ = 32 N - (102 N) sin 40^{\circ} = - 34 N$
(indicating that the force of friction is uphill) and to
$F_{N} - W_{A} cos θ = (102 N) c o s 40^{\circ} = 78 N$
which means that
$f_{s, m a x} = μ_{s} F_{N} = (0.56) (78 N) = 44$ N.
Since the magnitude f of the force of friction that holds the bodies motionless is less than $f_{s, m a x}$ the bodies remain at rest. The acceleration is zero.
(b) Since A is moving up the incline, the force of friction is downhill with
magnitude $f_{k} = μ_{k} F_{N}$ . Newton’s second law, using the same coordinates as in part (a),
leads to
$T - f_{k} - W_{A} sin θ = m_{A} a$
$F_{N} - W_{A} cos θ = 0$
$W_{B} - T = m_{B} a$
for the two bodies. We solve for the acceleration:
$a = \frac{W_{B} - W_{A} sin θ - μ_{k} W_{A} cos θ}{m_{B} + m_{A}} = \frac{32 N - (102 N) sin 40^{\circ} - (0.25) (102 N) c o s 40^{\circ}}{(32 N + 102 N) / (9.8 m / s^{2})}$
$= - 3.9 m / s^{2}$
The acceleration is down the plane, that is, $\to a = - (3.9 m / s^{2}) ι$ , which is to say (since the
initial velocity was uphill) that the objects are slowing down. We note that m = W/g has
been used to calculate the masses in the calculation above.
(c) Now body A is initially moving down the plane, so the force of friction is uphill with
magnitude $f_{k} = μ_{k} F_{N}$ . The force equations become
$T + f_{k} - W_{A} sin θ = m_{A} a$
$F_{N} - W_{A} cos θ = 0$
$W_{B} - T = m_{B} a$
which we solve to obtain
$a = \frac{W_{B} - W_{A} sin θ + μ_{k} W_{A} cos θ}{m_{B} + m_{A}} = \frac{32 N - (102 N) sin 40^{\circ} + (0.25) (102 N) c o s 40^{\circ}}{(32 N + 102 N) / (9.8 m / s^{2})}$
$= - 1.0 m / s^{2}$
The acceleration is again downhill the plane, that is, $\to a = - (1.0 m / s^{2}) ι$ . In this case, the
objects are speeding up.

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