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 0.35 and 0.25, respectively.
The question is incomplete, probably the figure has not been described.
I can visualize a box on a horizontal floor, but I do not see how and why the box would slip without any external force.
Please provide a complete description of the figure.
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 0.40 and 0.30, respectively.
To determine the minimum downward force required to prevent the box from slipping, we need to consider the forces acting on the box.
The static friction force acts in the direction opposite to the impending motion and is given by the equation:
F(static friction) = static friction coefficient * Normal force
The normal force is equal to the weight of the box, which can be calculated using the equation:
Weight (W) = mass (m) * acceleration due to gravity (g)
We can substitute the weight into the equation for the normal force:
Normal force = Weight = m * g
Now, the downward force is the sum of the weight and the static friction force, which are equal to the minimum downward force required to prevent slipping.
Therefore, the equation for the minimum downward force is:
Minimum Downward Force = Weight + F(static friction)
Minimum Downward Force = m * g + (coeff. of static friction * m * g)
Minimum Downward Force = m * g * (1 + coeff. of static friction)
Substituting the given values, we have:
Minimum Downward Force = m * 9.8 * (1 + 0.35)
Minimum Downward Force = m * 9.8 * 1.35
Hence, the minimum downward force on the box is 1.35 times the weight of the box.
To determine the minimum downward force required to prevent the box from slipping, we need to consider the forces acting on the box and the condition for equilibrium.
1. Draw a free-body diagram: Begin by drawing a diagram of the box and its forces. In this case, the downward force on the box is due to its weight (mg), and the two forces acting horizontally are the static friction force (fs) and the applied force (F).
---------------
| |
| Box |
| |
---------------
↑
mg
↓
-------------
← fs → F
-------------
2. Apply the condition for equilibrium: In order for the box to remain stationary (not slipping), the sum of the horizontal forces must be zero. This can be expressed as:
fs + F = 0
3. Determine the maximum static friction force: The static friction force can have a maximum value of fs_max = μs * N, where μs is the coefficient of static friction and N is the normal force. In this case, the normal force is equal to the weight of the box, so N = mg.
fs_max = μs * N = μs * mg
4. Substitute the values into the equation: We know that fs + F = 0 and fs_max = μs * mg, so we can rewrite the equation as:
μs * mg + F = 0
5. Solve for the downward force (F): To find the minimum downward force required to prevent slipping, rearrange the equation to solve for F:
F = - μs * mg
Since the coefficient of static friction given is 0.35, we substitute this value into the equation:
F = - 0.35 * mg
The negative sign indicates that the applied force must be in the opposite direction to counteract the static friction.
Therefore, the minimum downward force required on the box to prevent slipping is F = - 0.35 * mg.