One end of a post weighing 400 N and with height h rests on a rough horizontal surface with Ks=0.30. The upper end is held by a rope fastened to the surface and making an angle of 36.9^\circ with the post. A horizontal force F_vec is exerted on the post as shown. (Figure 1)
How large can the force be without causing the post to slip if its point of application is 6/10 of the way from the ground to the top of the post?
To find out how large the force can be without causing the post to slip, we need to consider the forces acting on the post and apply the conditions for equilibrium.
Let's start by identifying the forces acting on the post:
1. Weight of the post: The weight of the post acts vertically downward and has a magnitude of 400 N. We can represent this force as W = 400 N.
2. Normal force: The rough horizontal surface exerts an upward force on the post to counteract its weight. This force is perpendicular to the surface and can be denoted as N.
3. Friction force: Since the surface is rough, there will be a frictional force opposing the impending motion. We can represent this force as F_friction.
4. Applied force: The horizontal force F_vec exerted on the post, forming an angle of 36.9 degrees with the post.
In order to prevent the post from slipping, we need to ensure that the horizontal component of the applied force is less than or equal to the maximum possible friction force, which is given by F_friction = μ*N, where μ is the coefficient of static friction and N is the normal force.
Let's calculate the normal force:
Since the post is not accelerating in the vertical direction, the sum of the vertical forces must be zero. Therefore, N + (-W) = 0. Solving for N, we find N = W = 400 N.
Now, let's calculate the horizontal component of the applied force:
F_horizontal = F_vec * cos(36.9°)
To determine the maximum value of F_vec that would not cause the post to slip, it would need to be equal to or less than the maximum possible friction force.
F_horizontal ≤ F_friction
Substituting the values into the inequality, we have:
F_vec * cos(36.9°) ≤ μ * N
Now let's substitute the given values:
F_vec * cos(36.9°) ≤ 0.30 * 400 N
Simplifying further:
F_vec ≤ (0.30 * 400 N) / cos(36.9°)
Evaluate the right side of the inequality using a calculator:
F_vec ≤ 146.9 N
Therefore, the force F_vec can be at most 146.9 N without causing the post to slip.