A car weighing 1000 kg moves at a speed of 72 km / h on a concave bridge with a radius of 40 m. With what force does the car act on the bridge at its lowest point?
m = 1000 kg.
g = 9.8 m / s2.
V = 72 km / h = 20 m / s.
R = 40 m.
The force with which the body presses on the support or stretches the suspension is called the body weight P.
Two forces act on the car: the reaction force of the bridge N, directed vertically upwards;
the force of gravity m * g directed vertically downward.
Let’s write Newton’s 2 law for projections on the vertical axis directed vertically upward: m * a = – m * g + N.
N = m * a + m * g = m * (a + g).
Centripetal acceleration, and we will express it by the formula: a = V2 / R.
N = m * (V2 / R + g).
According to Newton’s 3 law, the force of action N is equal to the force of reaction P: P = N.
P = m * (V2 / R + g).
P = 1000 kg * ((20 m / s) 2/40 m + 9.8 m / s2) = 19800 N.
Answer: the weight of the car at the lowest points of the trajectory is P = 19800 N.
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