A box of mass m rests on a rough, horizontal surface with a coefficient of static friction μs. If a force F with arrowp is applied to the box at an angle θ as shown, what is the minimum value of θ for which the box will not move regardless of the magnitude of F with arrowp? (Use any variable or symbol stated above as necessary
To find the minimum value of θ for which the box will not move, we need to consider the forces acting on the box and apply the condition for static equilibrium.
Let's break down the forces acting on the box:
1. Weight (mg): This force pulls the box downward with a magnitude of mg, where m is the mass of the box and g is the acceleration due to gravity.
2. Normal force (N): This force is exerted by the surface and acts perpendicular to it. It balances the weight of the box and has a magnitude equal to the weight, N = mg.
3. Applied force (F): This force is applied at an angle θ and has a magnitude of F.
4. Frictional force (f): This force opposes the applied force and acts parallel to the surface. Its magnitude depends on the coefficient of static friction (μs) and the normal force (N).
To determine the minimum value of θ, we need to consider the limiting condition for static equilibrium, which means all the forces must balance each other. The condition can be expressed as:
ΣFhorizontal = 0
ΣFvertical = 0
In the horizontal direction, the applied force (F) and the frictional force (f) are the only forces present:
ΣFhorizontal = F - f = 0
In the vertical direction, the normal force (N) and the weight (mg) balance each other:
ΣFvertical = N - mg = 0
Since we want to find the minimum value of θ, we need to consider the maximum possible frictional force. The maximum frictional force can be found as:
fmax = μsN = μsmg
Substituting this value into the horizontal equilibrium equation:
F - fmax = 0
F - μsmg = 0
F = μsmg
Therefore, the minimum value of θ for which the box will not move is given by:
θ = arctan(F/mg)
θ = arctan(μs)
There is some frictional force that acts against the applied Force F to prevent the motion of the box.
F can be divided into its vertical component, Fsinθ, and its horizontal component, Fcosθ. The horizontal component is the one that acts against the frictional force and can cause the box to move.
As F goes on increasing, there will eventually come a time when Fcosθ is larger than the frictional force, for any non-zero value of cosθ.
But when θ is equal to 90 degrees, then there is no horizontal component, as Fcosθ is equal to zero. Hence, at this angle (when the force is directed into the ground), the box will not move regardless of the value of F, because cosθ is zero.