Answer:
1. [tex]W = w\sqrt{2}[/tex]
2. [tex]W = w\sqrt{2}/(1 +[/tex] μ[tex])[/tex]
Explanation:
This question is incomplete. Here is the complete question:
The leg press is an exercise machine on which the user pushes a platform away from their body using only their legs. The horizontal leg press has the platform connected to a stack of weights using a cable, with the force being translated directly without mechanical advantage. We will also assume that friction is negligible. The slanted leg press has the user sit near the floor and push a weighted sled up a track that is angled at [tex]45^0[/tex] from the floor. Arnold, a regular gym goer, can lift a maximum weight of [tex]w[/tex] on the horizontal leg press.
1. What is the maximum sled weight Arnold could press on the slanted machine (assuming zero friction)?
2. Arnold decides to try a different gym one day where he notices that the slanted leg press has not been well maintained. There is a coefficient of static friction μ between the track and sled. What is the maximum weight he can press on this machine?
From the question, we can see the different constraints at play as;
[tex]m[/tex] = mass of sled on slanted machine
[tex]g[/tex] = acceleration due to gravity = [tex]9.81m/s^2[/tex]
[tex]W[/tex] = weight of the sled = [tex]mg[/tex]
[tex]w[/tex] = Maximum weight that Arnold can lift on the leg press = Force ([tex]F[/tex])
1. Since the leg press being pushed by the user makes an angle of [tex]45^0[/tex] from the floor;
Force, [tex]F[/tex] = [tex]mg sin\alpha[/tex]
[tex]w = mg.sin\alpha\\w = W.sin\alpha\\[/tex]
[tex]\alpha = 45^0\\w = W({\frac{1}{\sqrt{2} }})\\W = w\sqrt{2}[/tex]
Therefore, on a slanted machine, Arnold can lift [tex]\sqrt{2}[/tex] times the weight which he can move on a horizontal leg press.
2. The friction force here can be obtained by the formula:
[tex]f =[/tex] μ[tex].mg.cos\alpha\\[/tex]
where μ = coefficient of static friction
The force now becomes:
[tex]F = mg.sin\alpha +[/tex] μ.[tex]mg.cos\alpha[/tex]
[tex]F = w = W(sin\alpha +[/tex] μ.[tex]cos\alpha)[/tex]
since
[tex]sin45^0 = cos45^0 = {\frac{1}{\sqrt{2} }}[/tex]
[tex]w = W({\frac{1}{\sqrt{2} }} +[/tex]μ[tex]/\sqrt{2})[/tex]
[tex]w = W(1+[/tex]μ[tex]/\sqrt{2})[/tex]
[tex]W = w\sqrt{2}/(1 +[/tex] μ[tex])[/tex]
