Question

3 forces in two dimensions (continued)
try it! (continued)

1. evaluate the answer

• are the units correct?
• is the magnitude of the velocity realistic?

check your progress

1. acceleration a rope pulls a 63-kg water skier up a 14.0° incline with a tension of 512 n parallel to the ramp. the coefficient of kinetic friction between the skier and the ramp is 0.27. what are the magnitude and direction of the skier’s acceleration?
2. forces one way to get a car unstuck is to tie one end of a strong rope to the car and the other end to a tree, then pull the rope at its midpoint at right angles to the rope. draw a free-body diagram and explain how even a small force on the rope can exert a large force on the car.

Explanation:

Problem 37 (Acceleration of Water Skier)

Step 1: Identify Forces Along Incline

Forces along the incline: Tension (\(T\)) up - friction (\(f_k\)) - component of weight (\(mg\sin\theta\)) down.
\(F_{net,x} = T - f_k - mg\sin\theta\)

Step 2: Calculate Normal Force (\(F_N\))

Perpendicular to incline: \(F_N = mg\cos\theta\)
\(m = 63\,\text{kg}\), \(g = 9.8\,\text{m/s}^2\), \(\theta = 14.0^\circ\)
\(F_N = 63 \times 9.8 \times \cos(14.0^\circ) \approx 63 \times 9.8 \times 0.9703 \approx 602.5\,\text{N}\)

Step 3: Calculate Kinetic Friction (\(f_k\))

\(f_k = \mu_k F_N\), \(\mu_k = 0.27\)
\(f_k = 0.27 \times 602.5 \approx 162.7\,\text{N}\)

Step 4: Calculate Net Force (\(F_{net,x}\))

\(T = 512\,\text{N}\), \(mg\sin\theta = 63 \times 9.8 \times \sin(14.0^\circ) \approx 63 \times 9.8 \times 0.2419 \approx 150.3\,\text{N}\)
\(F_{net,x} = 512 - 162.7 - 150.3 = 199\,\text{N}\) (up incline)

Step 5: Calculate Acceleration (\(a\))

\(F_{net} = ma \implies a = \frac{F_{net,x}}{m}\)
\(a = \frac{199}{63} \approx 3.16\,\text{m/s}^2\) (direction: up the incline)

Problem 38 (Forces to Unstuck Car)

Step 1: Free - Body Diagram (Rope Midpoint)

• Tension (\(T\)) from car (left) and tree (right), force (\(F\)) applied perpendicular (up).
• Resolve \(F\) into components: \(F = 2T\sin\theta\) (where \(\theta\) is angle between rope and horizontal).

Step 2: Analyze Force Magnitude

As \(\theta\) approaches \(90^\circ\), \(\sin\theta \approx 1\), but for small \(\theta\) (e.g., \(\theta \approx 1^\circ\)), \(\sin\theta \approx \theta\) (rad). Rearranging \(T=\frac{F}{2\sin\theta}\).
If \(\theta\) is small, \(\sin\theta\) is small, so \(T\) (force on car) is large even if \(F\) is small.

Answer:

s:

Problem 37:

Magnitude: \(\approx 3.16\,\text{m/s}^2\), Direction: Up the \(14.0^\circ\) incline.

Problem 38:

• Free - Body Diagram: Midpoint has two tension forces (along rope) and applied force (perpendicular).
• Explanation: Small \(F\) creates large \(T\) (force on car) because \(T=\frac{F}{2\sin\theta}\); small \(\sin\theta\) (small \(\theta\)) amplifies \(T\).