Posted by norm on Monday, March 15, 2010 at 9:21am.
What is the minimum downward force on the box in the figure that will keep it from slipping? The coefficients of static and kinetic friction between the box and the floor are 0.30 and 0.20, respectively. Box weighs 30 kg with 125 newtons pulling on horizontal surface.
The required friction force will depend upon the static coefficient. The required downward force Fy to make the friction force equal the pulling force Fx is given by
(Fy + Mg)*0.30 = Fx
Solve for Fy. M = 30 kg and Fx = 125 N
The kinetic friction coefficient is not needed.
To find the minimum downward force on the box that will keep it from slipping, we need to consider the frictional forces acting on the box.
The coefficient of static friction between the box and the floor is given as 0.30. This means that the maximum static frictional force that can act on the box is equal to the coefficient of static friction multiplied by the normal force.
The normal force acting on the box is equal to the weight of the box, which is given as 30 kg. The weight can be calculated using the equation W = mg, where W is the weight, m is the mass, and g is the acceleration due to gravity (approximately 9.8 m/s^2). So, the weight of the box is 30 kg * 9.8 m/s^2 = 294 N.
Therefore, the maximum static frictional force is 0.30 * 294 N = 88.2 N.
Since the box is being pulled horizontally with a force of 125 N, the maximum static frictional force of 88.2 N is greater than the pulling force. This means that the box will not slip as long as the pulling force is within the maximum static frictional force.
However, if the pulling force exceeds the maximum static frictional force, the box will start slipping. In that case, the box will experience kinetic friction, which has a coefficient of kinetic friction of 0.20.
To find the minimum downward force on the box that will keep it from slipping, we need to consider the kinetic frictional force.
The kinetic frictional force can be calculated using the equation Fk = μk * N, where Fk is the kinetic frictional force, μk is the coefficient of kinetic friction, and N is the normal force. Since the box is no longer experiencing static friction, the normal force remains the same at 294 N.
Therefore, the kinetic frictional force is 0.20 * 294 N = 58.8 N.
To prevent the box from slipping, the pulling force must be less than or equal to the kinetic frictional force. In this case, the pulling force is given as 125 N, which is greater than the kinetic frictional force of 58.8 N.
So, the minimum downward force on the box that will keep it from slipping is 58.8 N.
To summarize:
- If the pulling force is less than or equal to 88.2 N, the box will not slip and the static frictional force will keep it in place.
- If the pulling force is greater than 88.2 N but less than or equal to 125 N, the box will start slipping and experience kinetic friction.
- If the pulling force exceeds 125 N, the box will continue to slip.
These calculations are based on the information provided in the question. Make sure to double-check the values and units given in the original problem.