the coefficient of friction between a block of weight 15N and a horizontal board on which it is resting is 3.0find the horizontal force which will just remove the block and the force acting at an angle of 30 with the horizontal which will just move the block.
In the first case, the horizontal force needed to move the block is the (static) friction coefficient multiplied by the weight. That would be 45.0 N.
In the second case, the required friction force F is given by the formula
(Weight - Fsin30)*mu = F cos30
F[sin30*mu + cos30]= 15*mu
F = 15*mu/[1.5 + 0.866] = 19.0 N
To solve this problem, we can use the concepts of equilibrium and the relationships between forces in various directions. Let's break down the problem into two parts:
1. The horizontal force required to just remove the block:
In this case, we need to find the force that will overcome the friction and move the block. We can use the equation:
F_friction = μ * F_normal
where F_friction is the force of friction, μ is the coefficient of friction, and F_normal is the normal force (equal to the weight of the block). The normal force is given as 15N.
Given that the coefficient of friction is 3.0, we can substitute these values into the equation:
F_friction = 3.0 * 15N = 45N
Therefore, the horizontal force required to just remove the block is 45N.
2. The force at an angle of 30 degrees with the horizontal that will just move the block:
In this case, we need to find the force that will overcome friction and also the vertical component of the force to move the block at an angle. This requires adding the forces in both the horizontal and vertical directions.
Let F be the force required to move the block at an angle of 30 degrees.
The horizontal component of the force is given by F * cos(30°), and the vertical component is given by F * sin(30°).
The force of friction is still calculated using the same equation: F_friction = μ * F_normal.
For the block to remain in equilibrium, the sum of all vertical forces must be zero. This means that the vertical component of the applied force must be equal to the weight of the block, which is 15N.
Therefore, we have the following equations:
F_friction = μ * F_normal
F_friction = 3.0 * 15N = 45N
Vertical component: F * sin(30°) = 15N
From here, we can solve for F:
F * sin(30°) = 15N
F = 15N / sin(30°)
F ≈ 30N
Therefore, the force at an angle of 30 degrees with the horizontal that will just move the block is approximately 30N.