A 10 N horizontal force pushes a block weighing 4.9 N against a vertical wall (Fig. 6-26). The coefficient of static friction between the wall and the block is 0.55, and the coefficient of kinetic friction is 0.35. Assume that the block is not moving initially. In unit-vector notation (Fx i + Fy j), what is the force on the block from the wall?

Well, let's get to the bottom of this frictional force! Since the block is not initially moving, we are dealing with static friction. The force of static friction is given by fs ≤ μs * N, where fs is the force of static friction, μs is the coefficient of static friction, and N is the normal force.

To find the force on the block from the wall, we need to analyze the forces acting on the block. Since it is being pushed horizontally against the wall, the only force acting in the x-direction is the horizontal force of 10 N.

In the y-direction, we have the weight of the block (4.9 N) which acts downwards and the normal force (N) which acts upwards. Since the block is stationary, these forces must be equal in magnitude but opposite in direction. Therefore, N = 4.9 N.

Now, let's plug in the values. The force of static friction (fs) between the wall and the block is given by fs ≤ μs * N. By substituting the given values, we find that fs ≤ 0.55 * 4.9 N.

So, in unit-vector notation, the force on the block from the wall is (0.55 * 4.9 N) in the negative x-direction, which can be written as (-2.695 N i).

To find the force on the block from the wall, we need to consider the equilibrium conditions. Since the block is not moving initially, the sum of the forces in the horizontal direction (x-direction) must be zero.

Given:
Horizontal force (Fx) = 10 N (acting to the right)
Weight of the block (mg) = 4.9 N

The applied force (10 N) is counteracted by the frictional force, which is a combination of static and kinetic friction with the wall. Assuming no vertical forces, the frictional force acts in the negative x-direction.

The static friction force (fs) can hold the block in place until the force applied exceeds the maximum static friction force. The maximum static friction force can be found by multiplying the coefficient of static friction (μs) by the normal force.

Maximum static friction force (fstatic_max) = μs * (mg)

fstatic_max = 0.55 * (4.9 N)

Now, we can compare the applied force with the maximum static friction force to determine whether the block will move or not. If the applied force is less than or equal to the maximum static friction force, the block will not move, and the static friction force will balance the applied force.

In this case, the applied force (10 N) is greater than the maximum static friction force. So, the static friction force will reach its maximum value, trying to hold the block in place.

Therefore, the static friction force (fs) is equal to the applied force (10 N) and acts in the opposite direction (-x), i.e., fs = -10 N i.

Hence, the force on the block from the wall in unit-vector notation is:

Force on the block (F) = fs = -10 N i.

Note: The vertical force (Fy) is not considered in this situation as the block is not moving vertically, and there is no vertical acceleration.