A 50.0-kg box is resting on a horizontal floor. A force of 250 N directed at an angle of 26.0° below the horizontal is applied to the box. What is the minimum coefficient of static friction between the box and the surface required for the box to remain stationary?
1) 0.441 2) 0.654 3) 0.375 4) 0.866 5) 0.406
friction force=pushing force
mu(mg+250SinTheta)=250cosTheta
solve for mu
0.441
To solve this problem, we need to first find the force of static friction required to keep the box stationary.
The force of static friction can be calculated using the formula:
Fs = μs * N
where
Fs is the force of static friction,
μs is the coefficient of static friction, and
N is the normal force between the box and the surface.
The normal force can be calculated using the formula:
N = mg
where
m is the mass of the box (50.0 kg) and
g is the acceleration due to gravity (approximately 9.8 m/s^2).
N = (50.0 kg) * (9.8 m/s^2) = 490 N
Now, we can substitute the value of N into the equation for Fs:
Fs = μs * N
Given that the applied force is 250 N at an angle of 26.0° below the horizontal, we can find the vertical component of the applied force by multiplying the force by the sin of the angle:
Fy = 250 N * sin(26.0°)
Using the fact that the box is at rest, the force of static friction must be equal to the vertical component of the applied force, in order to keep the box stationary:
Fs = Fy
Substituting the values, we have:
μs * N = Fy
μs * 490 N = 250 N * sin(26.0°)
Now we can solve for the coefficient of static friction, μs:
μs = (250 N * sin(26.0°)) / 490 N
μs ≈ 0.441
Therefore, the minimum coefficient of static friction required for the box to remain stationary is approximately 0.441.
So the correct answer is option 1) 0.441.
To solve this problem, we can use Newton's second law of motion, which states that the net force acting on an object is equal to the product of its mass and acceleration:
ΣF = ma
In this case, the net force acting on the box is the horizontal component of the applied force, which can be found using trigonometry:
F_horizontal = F_applied * cos(θ)
where F_applied is the applied force and θ is the angle at which it is applied.
Next, we need to find the maximum static frictional force that can be exerted by the surface to keep the box stationary. The maximum static frictional force is given by:
f_static_max = μ_static * N
where μ_static is the coefficient of static friction and N is the normal force acting on the box.
The normal force N is equal in magnitude but opposite in direction to the gravitational force acting on the box:
N = mg
where m is the mass of the box and g is the acceleration due to gravity.
Now, let's calculate the horizontal component of the applied force:
F_horizontal = 250 N * cos(26.0°)
= 226.22 N
Next, let's calculate the maximum static frictional force:
f_static_max = μ_static * N
= μ_static * mg
Finally, for the box to remain stationary, the maximum static frictional force must be greater than or equal to the horizontal component of the applied force:
f_static_max ≥ F_horizontal
Substituting the expressions for f_static_max and F_horizontal, we have:
μ_static * mg ≥ 226.22 N
Simplifying the equation, we can cancel out the mass m:
μ_static * g ≥ 4.5244 N/kg
Now, solve for μ_static:
μ_static ≥ 4.5244 N/kg / g
The value of g is approximately 9.8 m/s².
Lets plug in the values:
μ_static ≥ 4.5244 N/kg / 9.8 m/s²
μ_static ≥ 0.461
Therefore, the minimum coefficient of static friction required for the box to remain stationary is approximately 0.461.
None of the given options match this answer, so the correct answer is not provided.