A uniform 250N ladder rest against a perfectly smooth wall, making a 35 angle with the wall
1) Calculate normal reaction force that the wall and floor exert on the ladder.
2) Calculate friction force on the ladder at the floor.
If you are lazy to calculate, please tell me the steps on how to approach this question. thx.
1) Calculate normal reaction force that the wall and floor exert on the ladder.
2) Calculate friction force on the ladder at the floor.
If you are lazy to calculate, please tell me the steps on how to approach this question. thx.
-
Consider the ladder leaning towards right on the wall.
The forces acting on the Ladder are Normal at the floor toward top (N(f)) Normal at the wall acting towards left (N(w)) . Frictional force acting towards the right at the end of the ladder f .
force mg acting downwards from the centre
Now we will take torque due to all the forces about the floor . We will consider the Length of the Ladder to be L .
T (mg) = mg(L/2) cos theta (Ladder makes theta angle with the ground ) \
T(Normal at the wall) = N(wall) X L sin theta
Now the ladder is in Equilibrium so equate the force
T(mg) = T(Normal at the wall)
mg(L/2) cos theta = N(wall) x L sin theta
Cancel L from both sides
so now we get the Normal from the wall = mg/2 cot theta
from the question that theta is 90 - 35 = 55 degrees .
so the normal the wall will exert will be 250/2 X cot55 degrees) = 0.7002075382097
87.5259422762 N
From the floor it would be 250 N itself.
Now for the calculation of the frictional force take the torque due the all the forces at the midpoint of the ladder .
and after equating those (its a long and pretty simple procedure just taking the forces in clock wise and anticlockwise directions and equating them dont consider mg as a force in this case . )
you will end up with
N(wall) sin 35 = mg cos55(1-u)
You know N at wall the values of angles and then calculate u which the frictional coefficient . Multiply that by mg (250N) and you will get the Frictional force .
The forces acting on the Ladder are Normal at the floor toward top (N(f)) Normal at the wall acting towards left (N(w)) . Frictional force acting towards the right at the end of the ladder f .
force mg acting downwards from the centre
Now we will take torque due to all the forces about the floor . We will consider the Length of the Ladder to be L .
T (mg) = mg(L/2) cos theta (Ladder makes theta angle with the ground ) \
T(Normal at the wall) = N(wall) X L sin theta
Now the ladder is in Equilibrium so equate the force
T(mg) = T(Normal at the wall)
mg(L/2) cos theta = N(wall) x L sin theta
Cancel L from both sides
so now we get the Normal from the wall = mg/2 cot theta
from the question that theta is 90 - 35 = 55 degrees .
so the normal the wall will exert will be 250/2 X cot55 degrees) = 0.7002075382097
87.5259422762 N
From the floor it would be 250 N itself.
Now for the calculation of the frictional force take the torque due the all the forces at the midpoint of the ladder .
and after equating those (its a long and pretty simple procedure just taking the forces in clock wise and anticlockwise directions and equating them dont consider mg as a force in this case . )
you will end up with
N(wall) sin 35 = mg cos55(1-u)
You know N at wall the values of angles and then calculate u which the frictional coefficient . Multiply that by mg (250N) and you will get the Frictional force .