if the coefficient of static friction between a 55 kg box and the floor is 0.67 with what minimum horizontal force must it be pulled to get the to crate move
m * g * μ = 55 kg * 9.8 m/s^2 * 0.67 ... Newtons
M*g = 55*9.8 = 539 N.
Fs = u*Fn = 0.67 * 539 = 361 N. = Force of static friction.
F-Fs = M*a.
F - 361 = M*0
F = 361 N.
To find the minimum horizontal force required to move the box, we need to consider the concept of static friction.
The formula for static friction is given by:
fs ≤ μs * fn
Where:
- fs represents the force of static friction
- μs is the coefficient of static friction
- fn is the normal force acting on the object
To determine the normal force acting on the box, we need to consider that the weight of an object is the force exerted by gravity and can be calculated by:
Fg = m * g
Where:
- Fg represents the force of gravity (weight)
- m is the mass of the object
- g is the acceleration due to gravity (approximately 9.8 m/s^2 on Earth)
Now, we can calculate the normal force:
fn = Fg = m * g
= 55 kg * 9.8 m/s^2
= 539 N
Given that the coefficient of static friction is 0.67, the formula for static friction becomes:
fs ≤ 0.67 * 539 N
Therefore, the minimum horizontal force required to overcome static friction and move the box is:
F = fs = 0.67 * 539 N
≈ 361.13 N
So, a minimum horizontal force of approximately 361.13 N must be applied to move the crate.