A box of mass m rests on a rough, horizontal surface with a coefficient of static friction μs. If a force Fp arrow is applied to the box at an angle θ as shown, what is the minimum value of θ for which the box will not move regardless of the magnitude of Fp? (Use any variable or symbol stated above as necessary.) The arrow Fp point at the box from the top-left direction and the answer is not 90 degree.
So I assume the angle theta is above the horizontal.
downward force on surface=mg+Fp*sinTheta
friction force=mu( Fp*sinTheta+mg)
horizontal force=Fp*cosTheta
when it wont move, friction>horizontal
mu *Fp*sinTheta+mg*mu>Fp*costheta
Fp(mu*sinTheta-cosTheta)>=mg*mu
That is not easy to solve, however graphically it may be easier.
To find the minimum value of θ for which the box will not move, we need to consider the forces acting on the box. The force of gravity is acting vertically downwards with a magnitude of mg, where g is the acceleration due to gravity.
The force applied to the box, Fp, can be resolved into its horizontal and vertical components: Fp_x and Fp_y, respectively.
Since the box is not moving vertically, the vertical component of the force applied, Fp_y, must balance out the force of gravity.
Therefore, Fp_y = mg.
Next, we need to consider the horizontal forces. The force of static friction, Fs, acts in the opposite direction of the applied force to prevent the box from moving horizontally.
The maximum value of static friction can be determined using the equation Fs = μs * N, where N is the normal force exerted on the box by the surface. In this case, the normal force is equal to the weight of the box, which is mg.
Therefore, Fs = μs * mg.
Since the applied force Fp_x is in the horizontal direction and the frictional force Fs is acting in the opposite direction, we can write:
Fp_x - Fs = 0
Substituting the values of Fs and Fp_x, we get:
Fp_x - μs * mg = 0
Rearranging this equation, we have:
Fp_x = μs * mg
From the given diagram, we can see that Fp_x = Fp * cos(θ) and Fp_y = Fp * sin(θ).
Since we want to find the minimum value of θ, we need to determine the angle at which the horizontal force Fp_x is equal to the maximum static friction force μs * mg.
Therefore, μs * mg = Fp * cos(θ).
Simplifying this equation, we have:
θ = arccos(μs * g / Fp)
To find the minimum value of θ for which the box will not move, we need to consider the forces acting on the box. Let's break down the forces:
1. The force applied to the box, Fp, can be resolved into two components:
- Fp₁: The component of the force parallel to the surface.
- Fp₂: The component of the force perpendicular to the surface.
2. The weight of the box, mg, acts vertically downward.
3. The normal force, N, acts perpendicular to the surface of the box.
4. The frictional force, Fr, acts parallel to the surface and opposes the applied force.
In order for the box to remain stationary, the frictional force must be equal to or greater than the component of the applied force parallel to the surface (Fp₁). This can be expressed as:
Fr ≥ Fp₁
The frictional force Fr can be calculated using the equation:
Fr = μs * N
The normal force N is equal to the weight of the box, N = mg.
If we consider the forces in the y-direction, the equation becomes:
mg - N = 0
Solving for N, we find:
N = mg
Now, substituting N into the equation for Fr, we have:
Fr = μs * mg
Substituting this into the inequality, we get:
μs * mg ≥ Fp₁
To find the value of Fp₁, we need to consider the horizontal component of the applied force, Fp₂:
Fp₁ = Fp * cos(θ)
Therefore, the inequality becomes:
μs * mg ≥ Fp * cos(θ)
If Fp is the maximum force that can be applied, we can assume that Fp > 0 and cos(θ) > 0. Therefore, we can divide both sides of the inequality by cos(θ):
μs * mg / cos(θ) ≥ Fp
The minimum value of θ for which the box will not move regardless of the magnitude of Fp occurs when the left-hand side of the inequality is equal to Fp, giving us the final equation:
μs * mg / cos(θ) = Fp
To calculate the minimum value of θ, we rearrange the equation to solve for cos(θ):
cos(θ) = μs * mg / Fp
Finally, we take the inverse cosine (or arccos) of both sides to find θ:
θ = arccos(μs * mg / Fp)
This gives us the minimum value of θ for which the box will not move.