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?
1) set mv^2/r= mu*mg
solve for mu
2) the vertical component of tension reduces weight.
friciton= mu*(mg-75sin35)
horizontalforce= 75*cos35
set them equal, solve for mu.
3) I will be happy to critiqeu your thinking.
thanks for your time and help, but I do not know what mu stands for
mu: coefficent of friction.
I got 13 for number 1, am I wrong?
Sure, I can help you with these questions. To find the coefficient of friction in each case, we can use the concept of frictional force.
1) To find the coefficient of friction required to keep the car on the road when rounding a turn of radius 25m at a speed of 20m/s, we need to consider the centripetal force acting on the car. The centripetal force is provided by the friction between the car's tires and the road, and is given by the equation:
Fc = m * v^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 looking for the coefficient of friction, we need to rewrite the centripetal force equation in terms of the coefficient of friction:
Fc = μ * m * g
where μ is the coefficient of friction and g is the acceleration due to gravity.
Equating both equations for the centripetal force, we can solve for the coefficient of friction μ:
μ * m * g = m * v^2 / r
μ = (v^2) / (r * g)
Substituting the given values, the coefficient of friction μ is:
μ = (20^2) / (25 * 9.8) = 0.82
Therefore, a coefficient of friction of at least 0.82 is necessary to keep the car on the road when rounding a turn of radius 25m at a speed of 20m/s.
2) To find the minimum coefficient of friction required for a man to slide a piece of furniture weighing 750N along a floor, we need to consider the forces acting on the furniture. The force of friction opposing the motion is given by:
Ff = μ * N
where Ff is the force of friction, μ is the coefficient of friction, and N is the normal force acting on the furniture.
The normal force N can be calculated using the given force applied by the man and the angle of the rope:
N = Fapplied * cos(angle)
Substituting the given values, the normal force N is:
N = 75N * cos(35 degrees) = 61.4N
Since the furniture is not moving, the force of friction must be equal and opposite to the applied force. Therefore, we have:
Ff = 75N * sin(35 degrees)
Now, we can solve for μ:
μ * N = Ff
μ * 61.4N = 75N * sin(35 degrees)
μ = (75N * sin(35 degrees)) / 61.4N = 0.465
Therefore, the minimum coefficient of friction required for the man to slide the furniture is 0.465.
3) To find the coefficient of friction necessary to hold a block stationary on a plane inclined at an angle of 30 degrees, we need to consider the forces acting on the block. The gravitational force acting on the block down the incline is given by:
Fgrav = m * g * sin(angle)
The normal force N acting perpendicular to the plane can be calculated as:
N = m * g * cos(angle)
The force of friction Ff opposing the motion is given by:
Ff = μ * N = μ * m * g * cos(angle)
In this case, we want to find the coefficient of friction μ when the block is held stationary, which means the frictional force must be equal to the gravitational force:
μ * m * g * cos(angle) = m * g * sin(angle)
μ = sin(angle) / cos(angle)
Substituting the given angle, the coefficient of friction μ is:
μ = sin(30 degrees) / cos(30 degrees) = 0.577
Therefore, a coefficient of friction of at least 0.577 is necessary to hold the block stationary on a plane inclined at an angle of 30 degrees.