How Do You Calculate Forces Acting on a Hamster and a Wedge on a Scale?

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SUMMARY

The discussion focuses on calculating the forces acting on a 200 g hamster sitting on an 800 g wedge block, which is placed on a spring scale. When static friction is sufficient, the combined mass of 1000 g results in a scale reading of 9.8 N. The analysis involves using force body diagrams and equations of motion to determine the normal forces acting on both the hamster and the wedge. The final calculations confirm that the normal force on the block is -8.99 N when static friction is no longer present.

PREREQUISITES
  • Understanding of Newton's laws of motion
  • Familiarity with force body diagrams
  • Knowledge of static and kinetic friction
  • Basic principles of vector resolution in physics
NEXT STEPS
  • Study the principles of static and kinetic friction in detail
  • Learn how to construct and analyze force body diagrams
  • Explore the concept of normal force in inclined planes
  • Review Newton's second law and its applications in multi-body systems
USEFUL FOR

This discussion is beneficial for physics students, educators, and anyone interested in understanding the dynamics of forces in static and kinetic scenarios, particularly in relation to inclined surfaces and frictional forces.

bcjochim07
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Force Body Diagram Hamster-Please Help!

Homework Statement


A 200 g hamster sits on an 800 g wedge shaped block. The block, in turn rests on a spring scale.

a) Initially static friction is sufficient to keep the hamster from moving. In this case the hamster and the block are effectively a single 1000g mass and the scale should read 9.8 N. Show that this is the case by treating the hamster and the block as separate objects and analyzing the forces.


Homework Equations





The Attempt at a Solution


I am reviewing, and I know I did this problem correctly before, but now it's not working for me. Could someone please tell me what I am doing wrong

Hamster: I drew a force body diagram, tilting the x & y axes

Sum of forces on y = Force of Block on hamster - mgcos40=0

Force of block on hamster = (.200kg)(9.80)(cos40)

Block: For the block I didn't tilt the coordinates

Sum of forces on y= normal force - Force of hamster on block in y direction - Mg
= n - (.200kg)(9.80)(cos40)(cos40) - (.800kg)(9.80)
n= -8.99 N which is the answer for the second part of the question, when the static friction goes away.
 
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Hi bcjochim07! :smile:

You're not asked for the separate friction force at this stage, so just calculate R, the total reaction force (normal + friction).

That makes only two forces on the hamster … R and mg.

So R = … ?

Then use minus R for the block. :smile:
 
Ok I got it, I forgot that friction has an equal opposite force
 

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