The discussion centers on the calculation of work done using the formula W=F·d·cos(θ), where θ represents the angle between the force and the direction of displacement. Participants clarify that θ is not always the angle between the force and displacement directly; it can also be the angle between the force and the perpendicular to the displacement. For example, in a problem involving a 70 kg man moving up stairs, the angle given is with respect to the horizontal, necessitating the use of trigonometric properties to determine the correct angle for calculations. Understanding these nuances is crucial for accurate application of the work formula.
PREREQUISITESThis discussion is beneficial for physics students, educators, and anyone involved in mechanics, particularly those learning about work, energy, and force interactions in various scenarios.
problem like:Doc Al said:
Note that the angle given is not between F and d, but between F and the perpendicular to d. It's up to you to figure out the correct angle!Sculter said:
Again, the angle given is the angle of the stairs with the horizontal. What you need is the angle between the force (gravity) and the displacement.Sculter said:*a man of mass 70kg. goes upstaris of lenth 50m. calculate the work done if g=10m/s2
-answer W=Fd cosθ, 70x10x50 x cos30
Here they just give you the required angle, so you can just plug it in.Sculter said:
To use the work formula you need the angle between the force and the displacement. Sometimes you'll be given that angle directly; other times you'll have to figure out the correct angle to use.Sculter said:
You are welcome.Sculter said:
Start by drawing yourself a diagram. And use the properties of right triangles. (For example, the two angles of a right triangle are complementary. They add to 90 degrees.)Sculter said: