A uniform 160 N ladder rests against a perfectly smooth wall, making a 70 degree angle with the wall (ladder is 7 m long). (a) find the normal forces that the wall and the floor exert on the ladder. B) what is the friction force on the ladder at the floor?
I tried to solve the problem and by making a sketch and go from there...
attempt: m1g(mass x gravity of ladder)cos(70)(7/2 is the lenght of the ladder divided by 2 because is in the center of the ladder that is exerted)= FW(force of the wall)sin(70)(7)...
160cos(70)(7/2)=FWsin(70)(7)
FW=[160cos(70)(7/2)]/[sin(70)(7)
from here on i do not know why my teacher told me to conver it to tan and he cross the lenghts but i do understand the reason for that, the only problem is the tan...
IT WILL BE VERY APPRECIATED IF SOMEBODY CAN EXPLAIN THE REASON FOR THE TAN PART
Your equations seem to indicate that the 70 degrees is with the FLOOR, not the Wall, which makes more sense)
First, force up = force down
so force up at floor = 160
Call friction force at floor Fw
It is equal and opposite to horizontal force at top
Now moments about ladder at floor sum to zero
160 cos 70 (L/2)= Fw sin 70(L)
80 = Fw sin 70/cos 70
BUT sin/cos = tan
so Fw = 80/tan 70 = 29.1 Newtons
To understand the reason for using the tangent in this problem, let's consider the forces acting on the ladder.
When the ladder is at rest and in equilibrium, the sum of the forces in the vertical direction (perpendicular to the wall) and horizontal direction (parallel to the wall) must be zero.
(a) Normal Forces:
The normal force exerted by the wall on the ladder can be found by considering the vertical forces. In this case, we have the weight of the ladder and the normal force by the floor acting downward, and the normal force by the wall acting upward. Since the ladder is at rest vertically, the sum of these forces must be zero.
Let's break down the vertical forces:
Weight of the ladder = mass x gravitational acceleration = m1g
Normal force by the floor = upward force = N
Normal force by the wall = upward force = N'
Since the ladder is in equilibrium vertically:
N + N' - m1g = 0
(b) Friction Force:
The friction force on the ladder at the floor can be found by considering the horizontal forces. In this case, we have the friction force acting in the opposite direction to the horizontal component of the normal force by the wall.
Let's break down the horizontal forces:
Friction force at the floor = opposing force = f
Horizontal component of the normal force by the wall = N'cos(70)
Since the ladder is in equilibrium horizontally:
f - N'cos(70) = 0
Now, let's analyze the reason for using tangent (tan) in the problem. The tangent is defined as the ratio of the opposite side to the adjacent side of a right-angled triangle.
In the given problem, we can form a right-angled triangle using the vertical and horizontal components of the normal force by the wall.
Opposite side = N' sin(70)
Adjacent side = N' cos(70)
Applying the tangent formula:
tan(70) = Opposite side / Adjacent side
tan(70) = N' sin(70) / N' cos(70)
N' sin(70) is the opposite side and N' cos(70) is the adjacent side.
By canceling out N', we get:
tan(70) = sin(70) / cos(70)
Hence, we use tangent to obtain the ratio of sin(70) to cos(70) in order to find the value of N'.