Question

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1. acceleration a rope pulls a 63 - kg water skier up a 14.0° incline with a tension of 512 n. the coefficient of kinetic friction between the skier and the ramp is 0.27. what are the magnitude and direction of the skiers 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. explain how even a small force on the rope can exert a large force on the car.
3. mass a large scoreboard is suspended from the ceiling of a sports arena by ten strong cables. six of the cables make an angle of 8.0° with the verticals and the other four make an angle of 10.0°. if the tension in each cable is 1300 n, what is the scoreboards mass?
4. equilibrium you are hanging a painting using two lengths of wire. the wires will break if the force is too great. should you hang the painting as shown on the left or the right of figure 20? explain.

figure 20

Explanation:

Problem 37

Step1: Define axes and forces

Let x = up incline, y = perpendicular to incline.
Forces: Tension \( T = 512\,\text{N} \), weight \( mg = 63 \times 9.8 = 617.4\,\text{N} \), friction \( f_k = \mu_k N \), normal force \( N \).

Step2: Y-direction equilibrium

\( N = mg\cos\theta = 617.4 \times \cos(14^\circ) \approx 617.4 \times 0.9703 \approx 600.0\,\text{N} \)

Step3: Calculate friction

\( f_k = 0.27 \times 600.0 = 162.0\,\text{N} \)

Step4: X-direction net force

\( F_{\text{net}} = T - mg\sin\theta - f_k = 512 - (617.4 \times \sin14^\circ) - 162 \)
\( \sin14^\circ \approx 0.2419 \), so \( mg\sin\theta \approx 617.4 \times 0.2419 \approx 149.3\,\text{N} \)
\( F_{\text{net}} = 512 - 149.3 - 162 = 200.7\,\text{N} \)

Step5: Find acceleration

\( a = \frac{F_{\text{net}}}{m} = \frac{200.7}{63} \approx 3.2\,\text{m/s}^2 \)

Free-body diagram for the car: tension force from the rope (horizontal, toward the tree). For the rope midpoint: two tension forces (along rope segments to car/tree) and applied pull (perpendicular). Resolving tension into vertical/horizontal components: small perpendicular pull creates large horizontal tension components (since \( T = \frac{F}{2\sin\theta} \), where \( \theta \) is small, \( \sin\theta \approx \theta \), so \( T \) is large).

Step1: Vertical tension per cable

For 6 cables: \( T_{\text{vert1}} = 1300 \times \cos8^\circ \approx 1300 \times 0.9903 = 1287.4\,\text{N} \)
For 4 cables: \( T_{\text{vert2}} = 1300 \times \cos10^\circ \approx 1300 \times 0.9848 = 1279.2\,\text{N} \)

Step2: Total vertical force

\( F_{\text{total}} = 6 \times 1287.4 + 4 \times 1279.2 = 7724.4 + 5116.8 = 12841.2\,\text{N} \)

Step3: Find mass

\( m = \frac{F_{\text{total}}}{g} = \frac{12841.2}{9.8} \approx 1310\,\text{kg} \)

Answer:

Magnitude: \( 3.2\,\text{m/s}^2 \), Direction: up the incline

Problem 38