A uniform crate with a mass of 16.2 kg rests on a floor with a coefficient of static friction equal to 0.571. The crate is a uniform cube with sides 1.21 m in length. what is the minimum height where the force F can be applied so that the crate begins to tip before sliding ?
I know that the horizontal force applied to the top of the crate that initiate the tipping is 79.5 N
The minimum distance will be achieved if the force is the maximum possible value without slipping, which is F = M*g*mus = 90.65 N
If the crate tips instead, the application height of the force H is such that there is a torque balance with the weight acting at the center of mass.
M*g*(1.21/2) = F*H
H = M*g*(1.21/2)/(M*g*mus) = 1.21/(2*mus) = 1.06 m
To find the minimum height where the force F can be applied so that the crate begins to tip before sliding, we need to consider the factors involved.
When the crate is on the verge of tipping, the static friction force on the bottom of the crate will be at its maximum. The maximum static friction force can be calculated using the equation:
friction force (F_f) = coefficient of static friction (μ) * normal force (F_n)
The normal force (F_n) acting on the crate is equal to the weight of the crate, which can be calculated by multiplying the mass of the crate by the acceleration due to gravity (9.8 m/s²):
F_n = mass (m) * acceleration due to gravity (g)
In this case, the mass of the crate is 16.2 kg, so:
F_n = 16.2 kg * 9.8 m/s²
Next, we need to calculate the maximum static friction force:
F_f = μ * F_n
The coefficient of static friction is given as 0.571, so:
F_f = 0.571 * (16.2 kg * 9.8 m/s²)
Now, we can calculate the torque (τ) acting on the crate due to the force F applied at a certain height. For the crate to begin tipping, the torque has to exceed the torque required to keep the crate from sliding.
The torque is given by the equation:
τ = F * h
Where F is the force and h is the height at which the force is applied.
The torque required to keep the crate from sliding is equal to the product of the maximum static friction force and the length of one side of the cube:
τ_required = F_f * length of one side (L)
The length of one side of the cube is given as 1.21 m.
So, we have:
τ_required = (0.571 * (16.2 kg * 9.8 m/s²)) * 1.21 m
Since we know that the horizontal force applied to the top of the crate that initiates the tipping is 79.5 N, we can set the torque equal to the torque required to keep the crate from sliding:
F * h = (0.571 * (16.2 kg * 9.8 m/s²)) * 1.21 m
Solving for h, the minimum height at which the force F can be applied:
h = [(0.571 * (16.2 kg * 9.8 m/s²)) * 1.21 m] / 79.5 N
Calculating this expression will give you the minimum height where the force F can be applied so that the crate begins to tip before sliding.
To determine the minimum height at which the force F can be applied so that the crate starts tipping before sliding, we need to consider the conditions required for tipping and sliding to occur.
Condition for Tipping:
The condition for tipping to occur is when the torque due to the applied force F is greater than or equal to the torque opposing it. In this case, the opposing torque is the torque due to the friction force acting at the bottom of the crate.
Condition for Sliding:
The condition for sliding to occur is when the friction force is at its maximum value, given by the static friction coefficient (μs) multiplied by the normal force (N) acting on the crate.
Let's calculate the normal force and the torque opposing the applied force:
1. Calculate the normal force (N):
Since the crate is at rest on the floor, the normal force is equal to the weight of the crate, which is mg, where m is the mass and g is the acceleration due to gravity (9.8 m/s²).
m = 16.2 kg
g = 9.8 m/s²
N = mg
N = 16.2 kg * 9.8 m/s²
N = 158.76 N
2. Calculate the torque opposing the applied force (τ):
The torque opposing the applied force is due to the friction force acting at the bottom of the crate. It can be calculated as the product of the friction force (Ff) and the lever arm (d), which is the half of the crate's side length (s/2).
Friction force (Ff) = coefficient of static friction (μs) * normal force (N)
Ff = 0.571 * 158.76 N
Ff = 90.69 N
The lever arm (d) = s/2
s = 1.21 m
d = 1.21 m / 2
d = 0.605 m
τ = Ff * d
τ = 90.69 N * 0.605 m
τ = 54.968 N·m
Now, we need to find the minimum height at which the tipping force is reached. The torque due to the applied force F should be greater than or equal to the torque opposing it (τ). The torque due to the applied force is given by the product of the applied force (F) and the lever arm (h), which is the height at which the force is applied.
τ = F * h
54.968 N·m = 79.5 N * h
Solving for h:
h = 54.968 N·m / 79.5 N
h ≈ 0.691 m
Therefore, the minimum height at which the force F can be applied so that the crate begins to tip before sliding is approximately 0.691 meters.