A uniform board is leaning against a smooth vertical wall. Find torque

[leezak]

A uniform board is leaning against a smooth vertical wall. The board is at an angle above the horizontal ground. The coefficient of static friction between the ground and the lower end of the board is 0.350. Find the smallest value for the angle , such that the lower end of the board does not slide along the ground.

i drew a diagram and all the forces and then i found out that the sum of the y-axis forces are normal force= (m)(g)(sin(theta)) and that the sum of the x forces are (m)(g)(cos(theta))=normal force(.350) i then plugged the normal (m)(g)(sin(theta)) from the first equation into where the normal force is in the second equation to get (m)(g)cos(theta)=(m)(g)(sin(theta))(.6) then i canceled out the mg and solved for theta to get 70.71 degrees but that's the wrong answer... what am i doing wrong?

 

[WalterContrata]

Good approach, but I think you are a little off. If the ladder is not moving, the sum of the forces in any direction must be zero. In particular, the sum of the forces in the y direction must be zero. Which forces have components in the y direction, and what are their magnitudes?

Hope this helps.

 

[quicknote]

Your approach to finding the torque is correct, but it seems like you made a mistake in your calculations. Here is the correct solution:

First, let's define the variables:
- m = mass of the board
- g = gravitational acceleration (9.8 m/s^2)
- theta = angle between the board and the ground
- mu = coefficient of static friction

Now, let's look at the forces acting on the board:
- Weight of the board: mg (acting downward)
- Normal force from the wall: N (acting perpendicular to the wall)
- Normal force from the ground: N' (acting perpendicular to the ground)
- Frictional force from the ground: muN' (acting parallel to the ground)

Since the board is not moving, the sum of the forces in the x-direction must be zero:
N' - muN' = 0
N' = muN'

Now, let's look at the sum of the forces in the y-direction:
N + N' - mgcos(theta) = 0
N + muN' - mgcos(theta) = 0
Substituting N' = muN':
N + mu^2N' - mgcos(theta) = 0

We also know that the board is in equilibrium, meaning that the sum of the torques must be zero. The torque from the weight of the board is mgcos(theta) and the torque from the normal force from the wall is Nsin(theta). Therefore, the equation for the sum of the torques is:
mgcos(theta) - Nsin(theta) = 0

Substituting N = -muN' and N' = muN' into this equation:
mgcos(theta) + mu^2N'sin(theta) = 0

Now, we can solve for theta:
mgcos(theta) + mu^2N'sin(theta) = 0
mgcos(theta) + mu^2(muN')sin(theta) = 0
mgcos(theta) + mu^3N'sin(theta) = 0
mgcos(theta) + mu^3(muN')sin(theta) = 0
mgcos(theta) + mu^4N'sin(theta) = 0
mgcos(theta) = -mu^4N'sin(theta)
tan(theta) = -mgcos(theta)/mu^4N'
tan(theta) = -mg/(mu^