Engineering

Name
Lecturer
Course
Date
AssignmentForces acting on the plank
There two main effects which operate on a stationary body. They are Action and reaction forces. They can be illustrated in the figure below:-

1485900787400

4000500163195

Action Force Weight of the decorator
A B
164465019748538862001974850

1028700831852787650197485

1219200197485
Reaction Force
Weight of the Plank
The weight of the painter and the weight of the Plank will act downward hence action force, and the rope will offer reaction force equivalent to the action force Knight, Randall Dewey (p34)
For the body to be stationary, then the two forces acting on it must be equal and must act in the opposite direction. In other words, Newton’s third law of motion must be in play, and it states that for a body to be stationary, then for every action force there is an equal and opposite reaction force. The two conditions are:-

There must be two equal forces, that is action and reaction for
Secondly, the forces must act in the opposite direction (37)
Using Newton’s third law,
FA= FB
Where FA is Action Force
FB is Reaction Force
Force = Ma
Where: – M = Mass
A= acceleration due to gravity =9.81 m/s2
The Mass of the man = 70Kg
Force = 70* 9.

81
= 686.7N
Force for Plank = 35 * 9.81
= 343.35N
Total Action force = 686.7 + 343.35
= 1030.05N
Action and Reaction forces are equal. Therefore if an action force is 1030.05N, then the minimum Reaction on the rope must be 1030.05N.

The formula for calculating internal stress is given by:-
The 20m converted = 20*10 = 200

The result shows that the internal stress of the rope is 2.40 indicating the strength of the rope.

Reference
Knight, Randall Dewey. Physics for scientists and engineers. Pearson Higher Ed., 2017.