91Ó°ÊÓ

For a Free Body Diagram (FBD), we have three forces acting on the car: 1. Weight of the car (W): This force acts vertically downward and has a magnitude of \(1000~kg * 9.81~m/s^2 = 9810~N\). 2. Normal Force (N): This force acts perpendicular to the inclined plane (hill) and counteracts the component of the weight that is perpendicular to the hill. The angle between the normal force and the vertical direction is the same as the hill's angle, which is \(5^\circ\). 3. Tension (T): This force is applied by the rope and acts at a \(10^\circ\) angle with respect to the horizontal. Now, we can draw a FBD with these forces and required angles on the car.

To calculate the tension in the rope, we need to apply Newton's second law of motion separately for two axes: horizontal \((x)\) and perpendicular to the inclined plane \((y)\). Since the car is being towed at constant speed, the net force acting on it is zero, meaning that forces in each direction will cancel out. For the horizontal \((x)\) axis, only the horizontal components of the tension force and normal force are acting. Thus, we can write the equation as: \(T_x - N_x = 0\) On the other hand, the vertical \((y)\) axis has the weight of the car acting downward and the vertical components of tension and normal force acting upward. Thus, we can write the equation as: \(N_y + T_y - W = 0\) Now, we need to find all components of the forces acting on the car, in terms of angles, using trigonometric functions: \(N_x = N \sin(5^\circ)\), \(N_y = N\cos(5^\circ)\) \(T_x = T\cos(10^\circ)\), \(T_y = T\sin(10^\circ)\) Substituting these expressions into our two equations: \((I) \quad T\cos(10^\circ) - N\sin(5^\circ) = 0\) \((II) \quad N\cos(5^\circ) + T\sin(10^\circ) = 9810~N\) Since the car is on the inclined plane, the normal force must counteract the perpendicular component of the weight. Thus, we can solve for the normal force (N): \(N\cos(5^\circ) = W\sin(5^\circ)\), \(N = W\frac{\sin(5^\circ)}{\cos(5^\circ)} = 9810~N \frac{\sin(5^\circ)}{\cos(5^\circ)}\) Now, substituting this expression for N into equation \((I)\), we can solve for the tension (T): \(T\cos(10^\circ) = 9810~N \frac{\sin(5^\circ)}{\cos(5^\circ)}\sin(5^\circ)\) \(T = \frac{9810~N \sin^2(5^\circ)}{\cos(10^\circ)\cos(5^\circ)} \approx 2190.76~N\) Thus, the tension in the rope is approximately \(2190.76~N\).