A uniform ladder which is 10metres long and weighs 300newtons leans with its upper end against smooth vertical wall and its lower end on arough ground.the bottom of the ladder is 6metres from the base of the wall.a man weighing 700newtons stands on the ladder at a point 6metres above the ground.calculate the magnitude and direction of the forces exerted on the ladder by (a)THE WALL(B)THE GROUND
To calculate the magnitude and direction of the forces exerted on the ladder by the wall and the ground, we can use the principles of equilibrium.
(a) Force exerted by the wall:
Since the ladder is in equilibrium, the sum of all the vertical forces acting on it must be zero. At the point where the ladder touches the wall, the force exerted by the wall acts perpendicular to the surface of the wall and cancels out the force due to the weight of the ladder.
To calculate the magnitude of this force, we sum up all the vertical forces acting on the ladder. In this case, the vertical forces are the weight of the ladder and the vertical component of the force exerted by the man.
Weight of the ladder = 300 Newtons (acting downward)
Vertical component of the man's weight = 700 Newtons × sin(theta), where theta is the angle between the ladder and the ground.
Since the ladder is in equilibrium, the sum of these forces must be zero:
300 Newtons - 700 Newtons × sin(theta) = 0
Now, we can solve this equation for sin(theta):
700 Newtons × sin(theta) = 300 Newtons
sin(theta) = 300 Newtons / 700 Newtons
sin(theta) ≈ 0.4286
To find the angle (theta), we can take the inverse sine (arcsin) of 0.4286:
theta ≈ arcsin(0.4286)
theta ≈ 26.86 degrees
Therefore, the magnitude and direction of the force exerted by the wall on the ladder is approximately 300 Newtons in a direction perpendicular to the wall.
(b) Force exerted by the ground:
To calculate the magnitude of the force exerted by the ground on the ladder, we need to consider the horizontal forces acting on the ladder.
In equilibrium, the sum of all the horizontal forces acting on the ladder must be zero. The horizontal forces are the horizontal component of the force exerted by the man and the force exerted by the ground.
Horizontal component of the man's weight = 700 Newtons × cos(theta), where theta is the angle between the ladder and the ground.
Since the ladder is in equilibrium, the sum of these horizontal forces must be zero:
Horizontal force exerted by the ground - Horizontal component of the man's weight = 0
Let's substitute the values we have:
Horizontal force exerted by the ground - 700 Newtons × cos(theta) = 0
Now, we can solve this equation for the horizontal force exerted by the ground:
Horizontal force exerted by the ground = 700 Newtons × cos(theta)
Using the value of theta we found earlier (approximately 26.86 degrees):
Horizontal force exerted by the ground = 700 Newtons × cos(26.86 degrees)
Calculating this value gives us:
Horizontal force exerted by the ground ≈ 636.38 Newtons
Therefore, the magnitude of the force exerted by the ground on the ladder is approximately 636.38 Newtons in the horizontal direction.
To summarize:
(a) The magnitude of the force exerted by the wall on the ladder is approximately 300 Newtons in a direction perpendicular to the wall.
(b) The magnitude of the force exerted by the ground on the ladder is approximately 636.38 Newtons in the horizontal direction.