A bar weighing 0.5 kg is pressed against a vertical wall by a force of 10N, directed horizontally, the coefficient

A bar weighing 0.5 kg is pressed against a vertical wall by a force of 10N, directed horizontally, the coefficient of sliding friction between the bar and the glass is 0.4.What is the minimum force that must be applied to the bar vertically in order to evenly lift it vertically upward with an acceleration of 2 m / s ^ 2

To find the force under which the specified bar will begin an accelerated upward movement, we project all the forces onto the vertical axis: m * a = F – Ftr – Ft = F – μ * N – m * g = F – μ * Fpr – m * g , whence we express: F = m * a + μ * Fпр + m * g.

Data: m is the mass of the specified bar (m = 0.5 kg); a is the required acceleration of the bar (a = 2 m / s2); μ is the coefficient of friction between the glass and the bar (μ = 0.4); Fпр – pressing force (Fпр = 10 N); g – acceleration due to gravity (g ≈ 10 m / s2).

Calculation: F = m * a + μ * Fpr + m * g = 0.5 * 2 + 0.4 * 10 + 0.5 * 10 = 10 N.

Answer: A vertical force of 10 N. is required to the specified bar.