Solving Hard Problems in Physics: Finding Minimum Force and Acceleration

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SUMMARY

The discussion focuses on solving physics problems involving forces on inclined planes and multiple blocks on a frictionless surface. The first problem involves calculating the minimum force required to prevent a 3.00 kg crate from sliding down a 35-degree incline with a static friction coefficient of 0.300, resulting in a solution of 32.2 N. The second problem addresses the acceleration and resultant forces on three blocks (m1=2kg, m2=3kg, m3=4kg) when a horizontal force of 180N is applied to m1. Key equations for resolving forces include the components of gravitational force and the normal force.

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  • Understanding of static friction and its coefficient
  • Knowledge of Newton's laws of motion
  • Ability to resolve forces into components on inclined planes
  • Familiarity with basic kinematics and dynamics of multiple bodies
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  • Study the concept of static friction in detail, focusing on applications on inclined planes
  • Learn how to resolve forces in two dimensions, particularly on inclined surfaces
  • Explore the dynamics of systems of multiple bodies and their interactions
  • Practice solving problems involving forces and accelerations using free-body diagrams
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allblonde
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:cry: some other problems if anyone has any ideas

1. the coefficient of static friction between the 3.00 kg crate and the 35 degree incline is .300. What is the magnitude of the minimum force, F, that must be applied to the crate perpendicular to the incline to prevent the crate from sliding down the incline?

: so i know the answer is 32.2 N but I am not sure how to get it , do i need to resovle the vectors of gravity(weight)

2.Three blocks are in contact with each other on a frictionless horizantal surface. A horizantal force, F, is applied to m1 (the first box). For this problem, m1=2kg, m2=3kg, m3=4kg, and F= 180N to the right.
a. find the acceleration of the blocks.
b. find the resultant force on each block.
c. find the magnitude of the contact forces.
 
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i know:
force perpendicular (y) is the force applied*the sin of theta
force parallel (x) is the force applied*the cos of theta
the force due to gravity is the mass*gravity
Fn+Fp-Fg= 0 because the crate is not floating
Fn=-Fp+Fg
Fk=coefficient of friction*Fn
Force perpendicular-Fk=m(a of x)

this is easy to understand and solve when the crate is on a flat horizantal surface, but when i try to plug these things into a problem with an incline, nothing comes out right i don't know if I am resolving wrong or what
 
allblonde said:
i know:
force perpendicular (y) is the force applied*the sin of theta
force parallel (x) is the force applied*the cos of theta
the force due to gravity is the mass*gravity
Fn+Fp-Fg= 0 because the crate is not floating
Fn=-Fp+Fg
Fk=coefficient of friction*Fn
Force perpendicular-Fk=m(a of x)

I think you're forces are incorrect for the first problem. The vertical force is perpendicular to the surface of the incline so there should not be any [tex]\theta[/tex] dependence. I have for my summation of forces
[tex]F_y: -F+F_n-mg\cos \theta[/tex]
and
[tex]F_x: -mg\sin \theta +\mu_{s}F_n=0[/tex]

The second problem is just doing the same thing three times or think of this as a center of mass problem. Good luck!
 
Last edited:

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