Weight on an Incline, a Two-Dimensional Problem

Problema e solução em inglês, como no livro original.

The skier’s mass, including equipment, is 60.0 kg. (See Figure 5.36 (b).) (a) What is her acceleration if friction is negligible? (b) What is her acceleration if the frictional force is 45.0 N?

A skier is moving down a slope with an angle measure of twenty-five degrees to the ground. Force vectors are shown: N points upward from the skier, f points to the right of the skis, w points at a downward left angle from the skiers' feet, forming an angle with vector w that has a measure of twenty-five degrees. A free-body diagram illustrates the right triangle formed by these vectors.

Figure 5.36 Now use the diagram to help find the skier's acceleration if friction is negligible and if the frictional force is 45.0 N.

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Estratégia
The most convenient coordinate system for motion on an incline is one that has one coordinate parallel to the slope and one perpendicular to the slope. Remember that motions along perpendicular axes are independent. We use the symbol $⊥$ to mean perpendicular, and $| |$ to mean parallel.

The only external forces acting on the system are the skier’s weight, friction, and the normal force exerted by the ski slope, labeled $w$ , $f$ , and $N$ in the free-body diagram. $N$ is always perpendicular to the slope and $f$ is parallel to it. But $w$ is not in the direction of either axis, so we must break it down into components along the chosen axes. We define $w_{| |}$ to be the component of weight parallel to the slope and $w_{⊥}$ the component of weight perpendicular to the slope. Once this is done, we can consider the two separate problems of forces parallel to the slope and forces perpendicular to the slope.
Solução
The magnitude of the component of the weight parallel to the slope is $w_{| |} = w sin (25 °) = m g sin (25 °)$ , and the magnitude of the component of the weight perpendicular to the slope is $w_{⊥} = w cos (25 °) = m g cos (25 °).$
(a) Neglecting friction: Since the acceleration is parallel to the slope, we only need to consider forces parallel to the slope. Forces perpendicular to the slope add to zero, since there is no acceleration in that direction. The forces parallel to the slope are the amount of the skier’s weight parallel to the slope $w_{| |}$ and friction $f$ . Assuming no friction, by Newton’s second law the acceleration parallel to the slope is
$a_{| |} = \frac{F_{net ||}}{m},$
Where the net force parallel to the slope $F_{net ||} = w_{| |} = m g sin (25 °)$ , so that
$\begin{matrix} a_{| |} & = & \frac{F_{net ||}}{m} = \frac{m g sin (25 °)}{m} = g sin (25 °) \\ = & (9.80 {m/s}^{2}) (0.423) = 4.14 {m/s}^{2} \end{matrix}$
is the acceleration.

(b) Including friction: Here we now have a given value for friction, and we know its direction is parallel to the slope and it opposes motion between surfaces in contact. So the net external force is now

$F_{net | |} = w_{| |} - f,$
and substituting this into Newton’s second law, $a_{| |} = \frac{F_{net ||}}{m}$ gives
$a_{| |} = \frac{F_{net ||}}{m} = \frac{w_{| |} - f}{m} = \frac{m g sin (25 °) - f}{m} .$
We substitute known values to get

$a_{| |} = \frac{(60 .0 kg)(9 {.80 m/s}^{2})(0 .423) - 45 .0 N}{60 .0 kg},$
or

$a_{| |} = 3 {.39 m/s}^{2},$
which is the acceleration parallel to the incline when there is 45 N opposing friction.

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