A uniform rod AB is 1.2 m long and weighs 16N. It is suspended by strings AC and BD. A block P weighing 96N is attached at E, 0.30N from A. The tension force in the string BD has a magnitude.
My work:
Taking torque around AC string
clockwise=counterclockwise
block P + Board force down= BD tension
(0.3m * 96N)+ (.6m * 16N)= BD (1.2m)
38.4= BD(1.2m)
BD= 32N
An 800N man stands halfway up a 5.0m long ladder of negligible weight. The base of the ladder is 3.0m from the wall. Assuming that the wall-ladder contact is frictionless, the wall pushes against the ladder with a force magnitude of:
my work:
cos (theta) = 3m/5m
theta= 53.13 degrees
horizontal components
Ff (friction foce on bottom) - FN (normal force on top) = 0
Ff=FN
vertical components:
Torque taken from top point of ladder
clockwise= counterclockwise
(800N X 2.5X cos(53.1))= FN(5)
FN=240N
Thank you
Was i right for the first problem?
yes.
also from what you wrote....we don't need Cos. I'm sorry I'm confused.
You can do it with the cosine, it gets the same answer. I just figured the distances along the normals from the diagram. Using cosines, you get the same thing.
Torque= force*distance*cosineangle OR
Torque= force*distance perpendicular between the force and point.
An 800N man stands halfway up a 5.0m long ladder of negligible weight. The base of the ladder is 3.0m from the wall. Assuming that the wall-ladder contact is frictionless, the wall pushes against the ladder with a force magnitude of:
my work:
cos (theta) = 3m/5m
theta= 53.13 degrees
horizontal components
Ff (friction foce on bottom) - FN (normal force on top) = 0
Ff=FN
vertical components:
Torque taken from top point of ladder
clockwise= counterclockwise
(800N X 2.5X cos(53.1))= FN(5)
FN=240N
***No. FN*5 is not the torque, Fn*4 is the torque. You need the perpendicular distance. And what happened to friction? Doesnt friction provide a ccw torque?
Summing around top:
cw= ccw
Ff*4 + 800*1.5 = Fv*3
But Fv= 800 , so Ff= 800*1.5/4= 300, but since summing horizontal forces, Ff= Fn, then Fn= 300
check my thinking.
In your calculations, you made a mistake in calculating the torque. The torque equation you used is correct, but you made an error in selecting the correct distance.
Let's reevaluate the problem step by step:
Given: An 800N man stands halfway up a 5.0m long ladder of negligible weight. The base of the ladder is 3.0m from the wall.
1. First, find the angle of the ladder with respect to the ground.
Use cos(theta) = 3.0m / 5.0m to find theta.
theta = cos^(-1)(3.0/5.0) = 53.1 degrees
2. Next, find the vertical force (Fn) exerted by the wall on the ladder.
Taking the sum of torques about the top of the ladder:
Clockwise = Counterclockwise
(800N * 2.5m * cos(53.1 degrees)) - Fn * 5.0m = 0
Simplifying the equation:
2000N * cos(53.1 degrees) - 5Fn = 0
Fn = (2000N * cos(53.1 degrees)) / 5.0m
Fn = 240N
So, the vertical force exerted by the wall on the ladder is 240N.
3. Finally, we need to find the horizontal force (Ff, friction force) exerted by the ground on the ladder.
Since the wall-ladder contact is assumed to be frictionless, there is no friction force in this case. Therefore, the horizontal force exerted by the wall on the ladder is zero.
In conclusion, the wall pushes against the ladder with a force magnitude of 240N vertically and zero horizontally since there is no friction force in this case.