I can't solve this at all.

[aftershock]

while moving in, a new homeowner is pushing a box across the floor at constant velocity. the coefficient of kinetic friction between the box and the floor is 0.41. the pushing force is directed downward at an angle (theta) below the horizontal. when theta is greater than a certain value, it is not possible to move the box, no matter how large the pushing force is.

find that value of theta.


I tried this.

P= pushing force
x = theta
m = mass of box
g= 9.8 or accel of gravity

Pcosx = (Psinx +mg).41

no luck :(

 

[willem2]

P= pushing force
x = theta
m = mass of box
g= 9.8 or accel of gravity

Pcosx = (Psinx +mg).41

no luck :(

Well you're nearly there. If the pushing force is very large then Psinx >> mg and you can ignore
the weight of the box.

 

[kerimek]

I understand your frustration with being unable to solve this problem. However, it is important to approach it systematically and use the appropriate equations. Let's start by setting up the forces acting on the box. We have the pushing force, P, directed downward at an angle x below the horizontal, and the force of kinetic friction, Fk, acting in the opposite direction. The force of gravity, mg, is also acting downward.

Using Newton's Second Law, we can write the following equation:
ΣF = ma

In this case, the acceleration, a, is zero since the box is moving at a constant velocity. Therefore, we can set ΣF = 0.

Next, we need to break down the forces into their components along the x-axis and y-axis. Along the x-axis, we have:
ΣFx = Pcosx - Fk

Along the y-axis, we have:
ΣFy = Psinx + mg

Since ΣF = 0, we can set ΣFx = 0 and solve for Fk:
Fk = Pcosx

We can then substitute this into our equation for ΣFy and solve for x:
Psinx + mg = Pcosx
sinx + (mg/P) = cosx
x = arccos(sin x + (mg/P))

The critical value of x, where the box will not move, is when Fk = μN, where μ is the coefficient of kinetic friction and N is the normal force. In this case, the normal force is equal to mg, so we can rewrite this as:
Pcosx = μmg

Substituting this into our equation for x, we get:
x = arccos(sin x + (mg/P)) = arccos(sin x + μm)

Therefore, the critical value of x, or the minimum angle at which the box will not move, is given by:
x = arccos(sin x + μm)

To find the numerical value of x, we need to know the values of P, m, and μ. Once those values are known, we can solve for x using the equation above.

In conclusion, while this problem may seem daunting, it can be solved using Newton's Second Law and breaking down the forces acting on the box. I encourage you to revisit the problem