A large 62.0 kg board is propped at a 44 degree angle against the edge of a barn door that is 3.1m wide.How great a horizontal force must a person behind the door exert (at the edge) in order to open it? Assume that there is negligible friction between the door and the board but that the board is firmly set against the ground
The board applies a normal force to the barn door. Its magnitude can be obtained by setting the magnitude of the torque abouut the ground support point equal to zero.
You don't say if the 44 degree angle is with respect to the grounmd or the door. I will assume the ground. In that case,
62*g*L*cos44 = F*L*sin44
F = 62*g*cot44 = 62.9 N
At the same distance from the hinge (i.e., "the edge"), the force on the pooposite side that is necessary to move the door is equal to F.
the answer was 310 N . Do you know why ?
I get 629 N, having misread the exponent of 10 last time. I cannot explain the discrepancy with 310 N. Changing the 44 degree reference from horizontal to vertical changes cotangent to tangent does not make a large difference.
Note that L is the length of the board, which cancels out
The weight force of the board is applied at the middle, if it is uniform.
Then
62 g (L/2) cos44 = F L sin44
F = 31 g cot44 = 314.6 N
They may be rounding to 310 because there are only two siginificant figures in the input data.
I forgot the 1/2 factor in my first answer
To find the horizontal force required to open the door, we need to consider the equilibrium of forces acting on the board.
First, let's resolve the weight of the board into its components. The weight acts vertically downward and can be divided into two components: the parallel component (mg * sinθ) and the perpendicular component (mg * cosθ).
The perpendicular component (mg * cosθ) does not contribute to the horizontal force required to open the door since it acts perpendicular to the door.
The parallel component (mg * sinθ) creates the force that needs to be counteracted by the person pushing on the door in order to prevent the board from sliding down.
We can calculate the parallel component using the formula:
F_parallel = mg * sinθ
Given:
Mass of the board (m) = 62.0 kg
Angle of inclination (θ) = 44 degrees
The acceleration due to gravity (g) is approximately 9.8 m/s^2.
Now, let's calculate the parallel component:
F_parallel = 62.0 kg * 9.8 m/s^2 * sin(44 degrees)
Using a scientific calculator, we can find the value of sin(44 degrees):
sin(44 degrees) ≈ 0.6947
F_parallel ≈ 62.0 kg * 9.8 m/s^2 * 0.6947
F_parallel ≈ 427.26 N
Therefore, a person behind the door must exert a horizontal force of approximately 427.26 N to prevent the board from sliding down and open the door.