Find the acceleration a with which a body of mass m = 3.5 kg moves under the action of three mutually

Find the acceleration a with which a body of mass m = 3.5 kg moves under the action of three mutually perpendicular forces F1 = 2 N, F2 = 3 N, F3 = 6 N

The resultant forces applied to the body will be equal to the vector sum of these forces. The modulus of the resultant can be found as the length of the diagonal of a rectangular parallelepiped, the sides of which are equal to the moduli of the forces acting on the body. Let’s find the resultant using the Pythagorean theorem:

F2 = F1 ^ 2 + F2 ^ 2 + F3 ^ 2 = 2 ^ 2 + 3 ^ 2 + 6 ^ 2 = 49 H2;

F = 7 H;

Let’s find the acceleration of the body by expressing it from Newton’s second law:

F = m * a;

a = F / m = 7 / 3.5 = 2 m / s2.

Answer: a = 2 m / s2.