The uniform block of width W and height H in the figure below rests on an inclined plane in static equilibrium. The plane is slowly raised until the block begins to tip over. Assume the tipping occurs before the block starts to slide. Find an expression for the angle θ at which the block is on the verge of tipping. (Use any variable or symbol stated above as necessary.)
mg is downward. cg is at H/2, w/2. So the point where mg is directly over the lower corner, (draw a sketch). TanTheta=h/2/w/2=h/w
and theta is that tipping angle.
TanTheta=tan^-1=W/H
is that right?
Depends on how you labeled the angle, bottom or top. Recheck your sketch to makecertain you have the tipping angle the one you want.
To find the angle θ at which the block is on the verge of tipping, we need to consider the equilibrium condition for the block.
Let's start by analyzing the forces acting on the block. There are two types of forces involved:
1. Gravity force (mg): The weight of the block acts vertically downward and can be split into two components: one parallel to the inclined plane and one perpendicular to it.
- The component parallel to the plane is mg * sin(θ), where θ is the angle of inclination.
- The component perpendicular to the plane is mg * cos(θ).
2. Normal force (N): The surface of the inclined plane exerts a normal force perpendicular to the plane. Since the block is in equilibrium, the normal force counters the perpendicular component of the weight.
- The normal force N = mg * cos(θ).
Now, considering the forces involved, we have:
1. Sum of forces in the x direction:
- Due to static equilibrium, the sum of forces in the x direction is zero.
- There is only one force component acting in the x direction, which is mg * sin(θ).
- Therefore, mg * sin(θ) = 0.
2. Sum of forces in the y direction:
- The sum of forces in the y direction should also be zero for the block to be in equilibrium.
- The forces acting in the y direction are mg * cos(θ) acting downward and the normal force N acting upward.
- Therefore, mg * cos(θ) + N = 0.
By substituting the value of N from the second equation into the first equation, we can solve for θ:
mg * sin(θ) = 0 (from the first equation)
mg * cos(θ) + N = 0 (from the second equation)
mg * cos(θ) + mg * cos(θ) = 0 (substituting N = mg * cos(θ))
2mg * cos(θ) = 0
cos(θ) = 0
θ = arccos(0)
θ = 90°
Therefore, the angle at which the block is on the verge of tipping is 90 degrees.