Find the Range of Forces for Which Block Does Not Slide

AI Thread Summary
To determine the range of forces for which a 0.5 kg block on a 2 kg wedge does not slide, a free-body diagram is essential. The forces acting on the block include gravity, which can be calculated as mgsin(40 degrees), and the static friction, dependent on the normal force. The normal force is influenced by the wedge's movement and the gravitational component acting perpendicular to the incline. Understanding the relationship between these forces is crucial for calculating the applied force F that prevents sliding. Properly summing the forces in both axes will yield the necessary values for F.
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Homework Statement



A block of mass .5 kg is on a wedge of mass 2.00 kg. The inclined surface makes an angle of 40 degrees with the horizontal. The wedge is subjected to a horizontqal force F and slides on a frictionless horizontal surface. The coefficient of static friction between the block and the wedge is .6 Find the range of values of F for which the block does not slide.
 
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... did you start by making a free-body diagram?

Try that see if you can sum forces...
 
Wesleytf said:
... did you start by making a free-body diagram?

Try that see if you can sum forces...

I've tried with the axes being x down the incline and y perpendicular to the incline.

sum of the forces in the x for the block is zero. The forces down the incline is gravity, which is mgsin 40 and coefficient of static friction*normal. I'm not sure what the normal is, and how the force applied to the wedge affects it.
 
The normal force is what keeps the block on the wedge. The component of g that is along your new y-axis will be equal in magnitude to the normal force. does that help?
 

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