To determine the smallest value of the force F such that the 2.0-kg block will not slide down the wall, we need to consider the forces acting on the block and the conditions for static equilibrium.
The forces acting on the block are:
Weight (W): The force exerted by gravity on the block, given by W = m * g, where m is the mass of the block and g is the acceleration due to gravity (approximately 9.8 m/s²).
Normal force (N): The force exerted by the wall on the block perpendicular to the wall's surface. In this case, the normal force is equal in magnitude and opposite in direction to the weight of the block.
Static friction force (F_s): The force of static friction acts parallel to the wall's surface and prevents the block from sliding down. The maximum static friction force can be determined using the equation F_s ≤ μ_s * N, where μ_s is the coefficient of static friction between the block and the wall, and N is the normal force.
For the block to remain in equilibrium and not slide down the wall, the maximum static friction force should balance the component of the weight pulling the block downward. This can be expressed as:
F_s = F * cos(θ) = W * sin(θ)
Where F is the applied force parallel to the wall, θ is the angle between the wall and the vertical direction, and W is the weight of the block.
In this case, since the block is not sliding, the static friction force F_s will be at its maximum, given by μ_s * N.
Setting the maximum static friction force equal to the weight's component:
μ_s * N = W * sin(θ)
Since the normal force N is equal in magnitude to the weight W, we can substitute W for N:
μ_s * W = W * sin(θ)
μ_s = sin(θ)
Given that the coefficient of static friction μ_s is 0.2, we can find the angle θ using the inverse sine (arcsine) function:
θ = arcsin(0.2)
θ ≈ 11.5 degrees
Therefore, the smallest value of the force F that prevents the block from sliding down the wall is when the applied force F is equal to the maximum static friction force:
F = μ_s * N = μ_s * W = μ_s * m * g
F = 0.2 * 2.0 kg * 9.8 m/s²
F ≈ 3.92 N
Hence, the smallest value of the force F required to prevent the block from sliding down the wall is approximately 3.92 Newtons.