Is it easier to push or to pull
- Slides: 13
Is it easier to push or to pull a lawnmower or vacuum cleaner?
Free Body Diagram Forces to consider y-axis Handle Push/Pull x-axis x a a a Lawnmower mg Force of Friction (Ff) Force Normal (FN)
Resolve the “Push/Pull” Vector Push Px = +P cos(a) Py = -P sin(a) P Pull Px = -P cos(a) Py = +P sin(a) Push/Pull (P) a Px Py
Push - FBD Handle Py x a a a Px Lawnmower mg Force of Friction (Ff) Force Normal (FN)
Push – Newton’s nd 2 Law Sum of Forces in the x-direction = Px – Ff = P cos(a) – u. FN (1) Sum of Forces in the y-direction = FN – Py – mg = FN – P sin(a) - mg = 0 for a body at rest Solving for FN: Substitute (2) into (1) Set equal to 0 FN = mg +P sin(a) x (2) Sum x = P cos(a) – u(mg + P sin(a)) 0 = P cos(a) – u(mg + P sin(a)) P(cos(a) – u sin(a)) = umg Or Push = umg/(cos(a)-usin(a)) (3)
Pull - FBD Handle Py x a a a Px Lawnmower mg Force of Friction (Ff) Force Normal (FN)
Pull – Newton’s nd 2 Law Sum of Forces in the x-direction = Ff – Px = u. FN – P cos(a) (4) Sum of Forces in the y-direction = FN + Py – mg = FN + P sin(a) - mg = 0 for a body at rest (5) Solving for FN: Substitute (2) into (1) FN = mg - P sin(a) x Sum x = -P cos(a) + u(mg - P sin(a)) Lawnmower Set equal to 0 0 = -P cos(a) + u(mg - P sin(a)) P(cos(a) + u sin(a)) = umg
Summary Calculations From our earlier calculations, we recall that at equilibrium (just before the lawnmower moves) “Push” and “Pull” may be represented as shown below: Push = umg/(cos(a)-usin(a)) (3) Pull = umg/(cos(a)+usin(a)) (6) Now, we know that cos(a) and sin (a) are less than or equal to 1. 00. Also, u is less than 1. 00. As a result, the denominator of (3) will always be less than the denominator of (6). Therefore, the “Push” will always exceed “Pull”. The force required to push a lawnmower is great than the force required to pull it.
Backup Slides •
Free Body Diagram Forces to consider y-axis Push/Pull Handle x-axis x a a a Lawnmower mg Force of Friction (Ff) Force Normal (FN)
Sum of Forces in the x-direction = Px – Ff = P cos(a) – u. FN Push - FBD (1) Sum of Forces in the y-direction = FN – Py – mg = FN – P sin(a) – mg = 0 for a body at rest Solving for FN: FN = mg +P sin(a) (2) Handle Py Substitute (2) into (1) Sum x = P cos(a) – u(mg + P sin(a)) Set equal to 0 0 = P cos(a) – u(mg + P sin(a)) P(cos(a) – u sin(a)) = umg Or Push = umg/(cos(a)-usin(a)) x a a a Px Lawnmower mg Force of Friction (Ff) Force Normal (FN)
Sum of Forces in the x-direction = Ff - Px = u. FN - P cos(a) Pull - FBD (1) Sum of Forces in the y-direction = FN + Py – mg = FN + P sin(a) – mg = 0 for a body at rest Solving for FN: FN = mg - P sin(a) (2) Handle Py Substitute (2) into (1) Sum x = -P cos(a) + u(mg - P sin(a)) Set equal to 0 0 = -P cos(a) + u(mg - P sin(a)) P(cos(a) + u sin(a)) = umg Or Pull = umg/(cos(a)+usin(a)) x a a a Px Lawnmower mg Force of Friction (Ff) Force Normal (FN)
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