Maximum value of F - SIN Waterloo Contest

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SUMMARY

The discussion centers on calculating the maximum force F that Superman can exert on a large block of mass M (450 kg) without causing a smaller block of mass m (250 kg) to slip. The static friction coefficient (μ) is 0.5, and the angle (∅) of the cable tied to the small block is 60°. The key equations involved are F=ma, Ff=Fnxμ, and Fnet=F1+F2+F3. Participants are encouraged to derive the static force equations for both masses to find the correct answer among the provided options.

PREREQUISITES
  • Understanding of Newton's laws of motion
  • Knowledge of static friction and its coefficient
  • Ability to draw and interpret free body diagrams (FBDs)
  • Familiarity with trigonometric functions in physics
NEXT STEPS
  • Review the derivation of static force equations for multiple masses
  • Practice solving problems involving friction on inclined planes
  • Learn about free body diagram techniques for complex systems
  • Explore the effects of varying angles on static friction scenarios
USEFUL FOR

This discussion is beneficial for physics students, educators, and anyone interested in mechanics, particularly those studying forces and friction in multi-body systems.

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Homework Statement



Superman is regaining his strength by pulling with a force F on a large block of mass M resting on a frictionless floor as shown. A small block of mass m rests on the upper surface of the large block, and there is friction at that surface, with static coefficient μ. Lois Lane has tied the small block to the wall with a light strong cable sloping up at an angle ∅. Everything is shown on the sketch, and things that look horizontal are. What is the maximum value of F before something slips? (Answers in Newtons)

∅- 60°
m=250kg
M=450kg
μ=0.5

Possible answers;

a) 1230

b) 1094

c) 946

d) 829

e) 656

*Check attachment for a picture.

Homework Equations



F=ma, Ff=Fnxμ,Fnet=F1+F2+F3

The Attempt at a Solution



I drew the FBDs for both masses, but I do feel i did something wrong.

for mass m, I had Tcos60° upwards, Tsin60° to the right, Ff to the left, and Fg facing down.

for mass M, I had Fg facing down, Fn facing up, and Fnm [Normal force of mass m] also facing down.

That's as far as I got.

*The picture is upside down, sorry!
 

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I think you have the cos and sin swapped.
You've left out the horizontal forces on mass M.
Please write out the static force equations for the two directions one each mass. You know what Fg is for each mass, so don't write it in that form (especially as having two forces with the same name is confusing).
 

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