A 67.0 kg crate rests on a level floor at a shipping dock. The coefficients of static and kinetic friction are .800 and .460 respectively.
A) what horizontal pushing force is required to just start the crate moving?
B) What horizontal pushing force is required to slide the crate across the dock at a constant speed.
If anyone could please help, I would greatly appreciate it. THANK YOU!
Physics assistance needed.
To solve this problem, we will use the equations for friction and Newton's second law of motion.
A) To find the horizontal pushing force required to just start the crate moving (the maximum force of static friction), we can use the equation:
F_static = μ_s * N
First, we need to calculate the normal force exerted by the floor on the crate. The normal force is equal to the weight of the crate, which is given by:
N = m * g
where m is the mass of the crate (67.0 kg) and g is the acceleration due to gravity (approximately 9.8 m/s^2).
N = 67.0 kg * 9.8 m/s^2 = 656.6 N
Now, we can substitute the values into the equation for static friction:
F_static = 0.800 * 656.6 N = 525.3 N
Therefore, the horizontal pushing force required to just start the crate moving is 525.3 N.
B) To find the horizontal pushing force required to slide the crate across the dock at a constant speed (the force of kinetic friction), we can use the equation:
F_kinetic = μ_k * N
We already know the normal force (N = 656.6 N) from the calculation in part A. Now, we can substitute the values into the equation for kinetic friction:
F_kinetic = 0.460 * 656.6 N = 301.8 N
Therefore, the horizontal pushing force required to slide the crate across the dock at a constant speed is 301.8 N.
Remember to always double-check your calculations and be mindful of units to ensure accurate results.