An ox exerts a force of 7100N to the horizontal direction without slipping. If the ox has a weight of 8000N, what is the minimum coefficient of static friction?
mus*8000 > 7100
mus > 0.887
(The actual value may be higher, since no slipping occurred)
"mus" is my notation for the static coefficient of friction
Hmm, let me calculate that for you. Hold on, I need to put on my thinking (red) nose for this one. *honk honk* Okay, I'm ready.
To find the minimum coefficient of static friction, we need to compare the force exerted by the ox to the maximum frictional force that can be generated.
The maximum frictional force is given by the formula: Ffriction = μs * Fnormal
Where:
Ffriction is the frictional force
μs is the coefficient of static friction
Fnormal is the normal force
Since the ox is not slipping, the frictional force must equal the force exerted by the ox. Therefore:
Ffriction = Force exerted by the ox = 7100N
The normal force is equal to the weight of the ox, which is 8000N.
Substituting these values into our formula:
7100N = μs * 8000N
Solving for μs:
μs = 7100N / 8000N
μs = 0.8875
So, the minimum coefficient of static friction is approximately 0.8875. That's as close as we can get to clowning around with numbers!
To find the minimum coefficient of static friction, we need to consider the condition where the force exerted by the ox is equal to the maximum frictional force that can be exerted without slipping.
The force exerted by the ox horizontally is 7100 N, and its weight is 8000 N. The weight provides a vertical force downwards that helps increase the friction.
To find the maximum frictional force, we use the equation:
Frictional force (Ffriction) = Coefficient of static friction (μ) * Normal force (N)
The normal force (N) is the force exerted by the ground onto the ox, which is equal to its weight (8000 N) since it's not accelerating vertically.
Therefore, we can rewrite the equation as:
Ffriction = μ * 8000 N
Since we want to find the minimum coefficient of static friction, we need to find the value of μ when the ox is exerting the maximum force without slipping.
The maximum force provided by the ox horizontally (7100 N) is the frictional force, so we have:
7100 N = μ * 8000 N
Now we can solve for μ:
μ = 7100 N / 8000 N
μ ≈ 0.8875
Therefore, the minimum coefficient of static friction is approximately 0.8875.
To find the minimum coefficient of static friction, we need to consider the equilibrium of forces acting on the ox.
The weight of the ox is acting vertically downward with a force of 8000N. The ox exerts a force of 7100N horizontally. Since the ox is not slipping, the static frictional force opposes the horizontal force to prevent slipping.
The equilibrium equation in the horizontal direction is:
Static frictional force = Horizontal force applied
∴ Fs = 7100N
The maximum static frictional force (Fs(max)) can be calculated using the formula:
Fs(max) = coefficient of static friction × normal force
Since the ox is not slipping, the static frictional force is at its maximum value, which is given by Fs(max). The normal force is the weight of the ox, which is 8000N.
∴ Fs(max) = coefficient of static friction × 8000N
Since we are looking for the minimum coefficient of static friction, we know that the static frictional force is also at its minimum value. Hence, Fs = Fs(min).
∴ Fs(min) = 7100N = coefficient of static friction × 8000N
Simplifying,
coefficient of static friction = 7100N / 8000N
coefficient of static friction = 0.8875
Therefore, the minimum coefficient of static friction is approximately 0.8875.