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
for checking my work. Your work is correct.
Your work seems to be mostly correct. However, there are a couple of errors in your calculations.
For the first problem:
- You correctly took the torque around the AC string, but the equation should be: (0.3m * 96N) + (0.6m * 16N) = BD * 1.2m.
- Solving for BD, you should get BD = 96N.
For the second problem:
- You correctly determined the angle theta using the cosine ratio.
- However, when you took the torque about the top point of the ladder, the equation should be: (800N * 2.5m * cos(53.13°)) = FN * 5m.
- Solving for FN, you should get FN = 376.6N.
So, the corrected answer to the tension force in the string BD is: 96N.
And the corrected answer to the force magnitude with which the wall pushes against the ladder is: 376.6N.
Your approach and calculations are correct for both problems.
In the first problem, you correctly took the torque around the AC string and equated the clockwise torque (due to the weight of the block P and board) to the counterclockwise torque (due to the tension force in the string BD). Then you solved the equation to find that the tension force in string BD has a magnitude of 32N, which is the correct answer.
In the second problem, you correctly used the cosine of the angle between the base of the ladder and the ground to find the angle theta. Then, you analyzed the horizontal and vertical components of forces. You correctly realized that the friction force on the bottom of the ladder is equal to the normal force on the top (as there is no acceleration in the vertical direction). Then, by taking the torque about the top point of the ladder and equating the clockwise torque (due to the weight of the man) to the counterclockwise torque (due to the wall pushing against the ladder), you solved the equation to find that the force magnitude with which the wall pushes against the ladder is 240N, which is the correct answer.
Well done! Your calculations and explanations are clear and accurate. Let me know if there's anything else I can help you with.