A block of mass 0.4 kg rests on the inclined surface of a wedge of mass 1.7 kg. The wedge is acted upon by a horizontal force F~ and slides on a frictionless surface. The acceleration of gravity is 9.81 m/s2.
If the coefficient of static friction between the wedge and the block is 0.9 and the angle of the incline is 31◦, find the minimum value of F~ for which the block does not slip. Answer in units of N
~ is the vector sign
To find the minimum value of the horizontal force F~ required for the block to not slip on the inclined surface of the wedge, we need to consider the forces acting on the block.
The gravitational force (mg) acting vertically downward can be broken down into two components:
1. The component parallel to the incline, which is mg*sin(θ), where θ is the angle of the incline (31° in this case).
2. The component perpendicular to the incline, which is mg*cos(θ).
Additionally, there will be a normal force (N) acting perpendicular to the incline.
In order for the block not to slip, the force of static friction (fs) acting parallel to the incline should be equal to or greater than the component of the gravitational force acting parallel to the incline (mg*sin(θ)).
The force of static friction can be found using the equation: fs = μ*N, where μ is the coefficient of static friction.
Since the force F~ acting on the wedge is horizontal, it will create a force component in the vertical direction that needs to be accounted for. This vertical component of the force F~ is F~ * sin(θ).
Now, let's analyze the forces acting on the block in the horizontal direction:
1. The force of static friction (fs) acting in the opposite direction of F~ will be fs = μ*N.
2. The horizontal component of the gravitational force (mg*sin(θ)) acting in the opposite direction of F~.
3. The horizontal component of the force F~ acting in the same direction as F~.
Since the wedge is sliding on a frictionless surface, there are no horizontal forces acting on it. Therefore, the horizontal net force acting on the block will be equal to the mass of the block (m) times its acceleration (a).
Now, we can set up the equation based on the forces in the horizontal direction:
F~ - fs = m*a
Substituting the values we have:
F~ - μ*N = m*a
We also need to consider the forces acting in the vertical direction:
1. The vertical component of the gravitational force (mg*cos(θ)) acting downward.
2. The normal force (N) acting upward.
3. The vertical component of the force F~ (F~ * cos(θ)) acting downward.
Since the block is not moving vertically, the vertical net force acting on the block will be zero.
Based on this, we can set up the equation:
mg*cos(θ) + N - F~ * cos(θ) = 0
From this equation, we can solve for N:
N = F~ * cos(θ) - mg*cos(θ)
Substituting this value of N into the equation for the force of static friction:
fs = μ*(F~ * cos(θ) - mg*cos(θ))
Now, we need to find the minimum value of F~ for which the block does not slip. This occurs when the force of static friction fs is at its maximum value:
fs = μ*N
μ*(F~ * cos(θ) - mg*cos(θ)) = μ*(F~ * sin(θ))
Canceling out μ, we can solve for F~:
F~ * cos(θ) - mg*cos(θ) = F~ * sin(θ)
F~ * cos(θ) = mg*cos(θ) + F~ * sin(θ)
F~ * cos(θ) - F~ * sin(θ) = mg*cos(θ)
F~ * (cos(θ) - sin(θ)) = mg*cos(θ)
Finally, solving for F~:
F~ = (mg*cos(θ)) / (cos(θ) - sin(θ))
Substituting the given values:
F~ = (0.4 kg * 9.81 m/s^2 * cos(31°)) / (cos(31°) - sin(31°))
Calculating this expression will give you the minimum value of F~ in units of Newtons.