The last car weighing 11 tons is moving uphill with acceleration (2.9 m / s ^ 2 on a traction force if the slope = 30 degrees, and the coefficient of friction is 0.04.
The resultant force F acting on a body with mass m can be found according to Newton’s second law F = m ∙ a, where a is acceleration. On the other hand, the resultant force F is found as the vector sum of all forces acting on the body: traction forces – F thrust, friction forces – Ffr, support elastic (reaction) forces – N, gravity – Ftyazh. To find this sum, let’s connect the reference system to the car: we will place the reference point at its center of gravity, the abscissa axis will be directed along the trajectory of movement, the ordinate axis – perpendicularly upward. Let’s write down the coordinates of the forces in the selected frame of reference. The resultant force F has coordinates (m ∙ a; 0), the thrust Fthroat has coordinates (Fthrust; 0), the friction force Ffr has coordinates (- μ ∙ N; 0), the reaction forces of the support N has coordinates (0; N), of gravity Ftyazh has coordinates (- m ∙ g ∙ sinα; – m ∙ g ∙ cosα), where α is the slope angle, g = 9.8 m / sec².
Let us equate the abscissa of the resultant force m ∙ a to the sum of the abscissas of all forces, and the ordinate of the resultant force 0 to the sum of the ordinates of all forces, we obtain the system of equations:
m ∙ a = Fthrust – μ ∙ N + 0 – m ∙ g ∙ sinα and 0 = 0 + 0 + N – m ∙ g ∙ cosα.
Then N = m ∙ g ∙ cosα; Fthrust = m ∙ a + μ ∙ m ∙ g ∙ cosα + m ∙ g ∙ sinα. We get Fthrust = m ∙ (a + g ∙ (μ ∙ cosα + sinα)). From the condition of the problem it is known that a car with a mass of m = 11 tons = 11000 kg is moving uphill with an acceleration of a = 2.9 m / s² and an inclination of α = 30 degrees, and the coefficient of friction is μ = 0.04. Substitute the values of physical quantities in the formula and make calculations: Fthrust = 11000 ∙ (2.9 + 9.8 ∙ (0.04 ∙ cos30 ° + sin30 °)); Fthrust ≈ 89534 N.
Answer: the traction force is ~ 89,534 N.
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