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.
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
4 answers
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
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
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
2 answers