Max Breaking Force on Car & Friction of Mop: Coefficient, Mass, Angle

[me!ü]

The coefficient of static friction between the tires of a car and dry road is 0.62. The mass of the car is 1600 kg. What maximum breaking force is obtainable (a) on a level road and (b) 8.6° downgrade?

The handle of a floor mop of mass m makes an angle θ with the vertical direction. Let μk be the coefficient of kinetic friction between mop and floor and μs the coefficient of static friction between mop and floor. Neglect the mass of the handle. (a) Find the magnitude of the force F directed along the handle required to slide the mop with uniform velocity across the floor. (b) Show that θ is smaller than a certain θ0 of the mop cannot be made to slide across the floor no matter how great a force is directed along the handle. What is angle θ0?

 

[tiny-tim]

Hi me!ü!

Show us what you've tried, and where you're stuck, and then we'll know how to help!

 

[tomcue]

I would like to address the content provided by analyzing the given information and providing a response in a clear and concise manner.

Firstly, the coefficient of static friction between the tires of a car and dry road is a measure of the force required to keep the car from sliding on a level surface. In this case, the coefficient of static friction is 0.62. This means that the maximum breaking force that can be obtained on a level road is 0.62 times the weight of the car, which is equivalent to 0.62 x 1600 kg = 992 N.

On a downgrade of 8.6°, the maximum breaking force that can be obtained is equal to the weight of the car, which is 1600 kg x 9.8 m/s^2 = 15,680 N. This is because the force of gravity acting on the car increases as the angle of the downgrade increases.

Moving on to the second part of the content, we are given information about a floor mop and its coefficients of kinetic and static friction. The coefficient of kinetic friction, μk, is a measure of the force required to keep the mop moving across the floor at a constant velocity. The coefficient of static friction, μs, is the force required to prevent the mop from sliding across the floor.

(a) To find the magnitude of the force required to slide the mop with uniform velocity, we can use the formula F = μkmg, where m is the mass of the mop and g is the acceleration due to gravity (9.8 m/s^2). Neglecting the mass of the handle, the force F required to slide the mop would be equal to F = μkmg = μk x 0.5m x 9.8 m/s^2 = 4.9 μk N, where 0.5m is the center of mass of the mop.

(b) In order for the mop to slide across the floor, the angle θ must be smaller than a certain angle θ0. This is because as the angle increases, the force of gravity acting on the mop also increases, making it more difficult for it to slide. The value of θ0 can be found using the equation μk = tan θ0, where μk is the coefficient of kinetic friction. Therefore, θ0 = tan^-1 (μk) = tan^-1 (0.