Please help me with these three questions. They deal with coefficient of friction

1) A car rounds a turn of radius 25m at a speed of 20m/s. Find the coefficient of friction neccessary to keep the car on the road

2) A man is attempting to slide a piece of furniture weighing 750N long a floor. Ifhe exerts a force of 75N along a rope and angle of 35 degrees and the object does not move, what is the minimum value for te coeficient of friction?

3) Consider a block on a plane inclined at an angle of 30 degres. If the block weights 950N, what coefficient of friction is neccessary to hold it stationary?

Sure! Let's go through each question step by step:

1) To find the coefficient of friction necessary to keep the car on the road, we can use the concept of centripetal force. The centripetal force required to keep the car on a curved path is provided by the friction between the tires of the car and the road. Mathematically, it is given by the equation:

Fc = mv^2/r

where Fc is the centripetal force, m is the mass of the car, v is the velocity of the car, and r is the radius of the turn.

Since we are given the radius of the turn (25m) and the speed of the car (20m/s), we can rearrange the equation to solve for the coefficient of friction. The formula for friction force is:

Ff = μN

where Ff is the frictional force, μ is the coefficient of friction, and N is the normal force (equal to the weight of the car, mg).

In this case, the frictional force (Ff) is equal to the centripetal force (Fc). So we have:

μN = mv^2/r

Substitute N = mg:

μmg = mv^2/r

Now, divide both sides by mg:

μ = v^2/(rg)

Substituting the given values, we get:

μ = (20^2) / (25 * 9.8)

Simplifying and calculating, the coefficient of friction required to keep the car on the road is approximately 0.82.

2) In this question, we need to find the minimum value for the coefficient of friction so that the object does not move. The force applied by the man is 75N along a rope at an angle of 35 degrees. The vertical component of this force (Fv) contributes to countering the weight of the object, while the horizontal component (Fh) contributes to overcoming the frictional force. The equation for the horizontal component is:

Fh = F * cos(θ)

where F is the force applied by the man and θ is the angle of the rope with respect to the horizontal.

In this case, θ is 35 degrees, and F is given as 75N. So:

Fh = 75 * cos(35)

Now, the frictional force (Ff) is equal to the coefficient of friction (μ) multiplied by the normal force (N). In the absence of vertical motion, N is equal to the weight of the object (750N). So:

Ff = μN = μ * 750

Since the horizontal force (Fh) is trying to overcome the frictional force, we can write:

Fh ≥ Ff

Substituting the values:

75 * cos(35) ≥ μ * 750

Divide both sides by 750:

(75/750) * cos(35) ≥ μ

Simplifying and calculating, the minimum value for the coefficient of friction is approximately 0.104.

3) To find the coefficient of friction necessary to keep the block stationary on an inclined plane, we need to consider the forces acting on the block. The force of gravity (mg) is acting vertically downward, and the component of this force (mg sin(θ)) parallel to the incline contributes to the downhill force. The frictional force (Ff) acts in the opposite direction to resist the block from sliding.

The equation for the frictional force is:

Ff = μN

To find the normal force (N), we need to consider the components of the weight. The vertical component (mg cos(θ)) counteracts the normal force, so:

N = mg cos(θ)

Now, we can substitute this into the equation for frictional force:

Ff = μ * mg cos(θ)

Since we want to hold the block stationary, the frictional force should be equal to or greater than the downhill force. The downhill force is given by the component of gravity parallel to the incline, which is mg sin(θ). So we have:

μ * mg cos(θ) ≥ mg sin(θ)

Divide both sides by mg:

μ cos(θ) ≥ sin(θ)

Now, divide both sides by cos(θ):

μ ≥ tan(θ)

In this case, the angle (θ) is 30 degrees. Therefore, the coefficient of friction necessary to hold the block stationary is equal to the tangent of 30 degrees, which is approximately 0.577.