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 oor and push a weighted sled up a track that is angled at 45o from the oor. Arnold, a regular gym goer, can lift a maximum weight of w on the horizontal leg press.

1.

2. μ

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 from the floor. Arnold, a regular gym goer, can lift a maximum weight of 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;

= mass of sled on slanted machine

=  acceleration due to gravity =

=  weight of the sled =  

= Maximum weight that Arnold can lift on the leg press = Force ()

1. Since the leg press being pushed by the user makes an angle of from the floor;

Force, =

Therefore, on a slanted machine, Arnold can lift times the weight which he can move on a horizontal leg press.

2. The friction force here can be obtained by the formula:

μ

where μ =  coefficient of static friction

The force now becomes:

μ.

μ.

since

μ

μ

μ

0.19 m/s and 0.45 m/s

double the distance = divide strength by 4