A block of mass 9.00 kg is pushed up against a wall by a force P that makes a 50.0° angle with the horizontal as shown below. The coefficient of static friction between the block and the wall is 0.245. Determine the possible values for the magnitude of P that allow the block to remain stationary.
So you are pushing a block at an angle of 50° against a vertical wall, to keep it stationary what is Pmin force, and Pmax force (before the block starts moving upwards).
Pmin=?
Pmax=?
Force down = m g
Force up = F sin 50 + .245 F cos 50
so to keep it from falling
F (sin 50 + .245 cos 50) = m g
to start it upwards, reverse the friction force
F (sin 50 -.245 cos 50) = m g
Thank you!!
To keep the block stationary, the force pushing the block against the wall (P) must be equal to or less than the maximum static friction force between the block and the wall.
Let's calculate the maximum static friction force first:
1. Determine the normal force acting on the block. Since the block is pushed against the wall, the normal force (N) is equal to the weight of the block, given by N = mg, where m is the mass of the block and g is the acceleration due to gravity. In this case, m = 9.00 kg and g ≈ 9.8 m/s², so N = (9.00 kg)(9.8 m/s²) = 88.2 N.
2. Calculate the maximum static friction force. The maximum static friction force (f_friction) is given by the product of the coefficient of static friction (μ_s) and the normal force (N), i.e., f_friction = μ_sN. Here, μ_s = 0.245, so f_friction = (0.245)(88.2 N) = 21.59 N (approx.).
Now we can determine the possible values for the magnitude of P:
3. Resolving forces in the vertical direction. Since the block is stationary, the vertical component of the force P must be equal to the weight of the block, mg, to balance it:
P * sin(50°) = mg
4. Solving for P:
P = mg / sin(50°)
= (9.00 kg)(9.8 m/s²) / sin(50°)
≈ 113.08 N
Therefore, the minimum force (Pmin) required to keep the block stationary is approximately 113.08 N.
5. Resolving forces in the horizontal direction. The horizontal component of the force P must be less than or equal to the maximum static friction force, f_friction, to prevent the block from sliding up the wall:
P * cos(50°) ≤ f_friction
6. Solving for P:
P ≤ f_friction / cos(50°)
= 21.59 N / cos(50°)
≈ 29.89 N
Therefore, the maximum force (Pmax) that can be applied before the block starts moving upwards is approximately 29.89 N.
To determine the possible values for the magnitude of P that allow the block to remain stationary, we need to consider the forces acting on the block.
First, let's identify the forces acting on the block:
1. The gravitational force (mg) acting straight downward with a magnitude of (9.00 kg) x (9.8 m/s^2) = 88.2 N.
2. The normal force (N) exerted by the wall, which acts perpendicular to the wall. Since the block is stationary in the vertical direction, the normal force must balance the gravitational force, so N = mg = 88.2 N.
3. The force of static friction (fs) exerted by the wall, which acts parallel to the wall and opposes the motion of the block. The maximum static friction force can be determined using the formula fs(max) = μsN, where μs is the coefficient of static friction.
Since the block is at the verge of moving upward, the maximum static friction force fs(max) would be equal to the component of the pushing force P that acts parallel to the wall. The component of P parallel to the wall is Pcosθ, where θ is the angle between P and the horizontal direction.
Now, let's calculate the possible values for P:
Given: θ = 50° and μs = 0.245
1. Calculate the maximum static friction force:
fs(max) = μsN
fs(max) = (0.245)(88.2 N)
fs(max) ≈ 21.56 N
2. Since the maximum static friction force fs(max) equals the parallel component of P, we have:
fs(max) = Pcosθ
P = fs(max) / cosθ
P = 21.56 N / cos(50°)
P ≈ 33.75 N
Therefore, the possible values for the magnitude of P that allow the block to remain stationary are:
Pmin = 0 N (no force required to keep the block stationary)
Pmax ≈ 33.75 N