A box weighing 60 kg begins to be moved along a horizontal surface with an acceleration of 1 m / s ^ 2
A box weighing 60 kg begins to be moved along a horizontal surface with an acceleration of 1 m / s ^ 2, acting on it with a constant force directed at an angle of 30 degrees to the horizontal. Determine the force with which the box is pulled if the sliding friction coefficient is 0.2?
m = 60 kg.
g = 9.8 m / s ^ 2.
μ = 0.2.
a = 1 m / s ^ 2.
∠α = 30 °.
F -?
Let’s write 2 Newton’s law in vector form for a box: m * a = F + Ftr + N + m * g.
Let’s find expressions 2 of Newton’s law in projections on the axis.
ОХ: m * a = F * cosα – Ftr.
OU: 0 = F * sinα + N – m * g.
N = m * g – F * sinα.
The friction force is determined by the formula: Ffr = μ * N = μ * (m * g – F * sinα) = μ * m * g – μ * F * sinα.
m * a = F * cosα – μ * m * g + μ * F * sinα.
F * cosα + μ * F * sinα = m * a + μ * m * g.
F * (cosα + μ * sinα) = m * (a + μ * g).
F = m * (a + μ * g) / (cosα + μ * sinα).
F = 60 kg * (1 m / s ^ 2 + 0.2 * 9.8 m / s ^ 2) / (cos30 ° + 0.2 * sin30 °) = 185 N.
Answer: the box is pulled with a force of F = 185 N.