Finding Force via Static Friction

AI Thread Summary
To determine the force required to start moving a crate across a rough floor, the problem involves calculating the normal force and static friction. The crate has a mass of 34 kg and a coefficient of static friction of 0.48, with the applied force at a 21° angle below the horizontal. The normal force is calculated as N = mgcos(θ), resulting in approximately 311.386 N, while the maximum static friction force is fs,max = μs*N, yielding about 149.465 N. The necessary force to initiate movement is derived using the equation F = μs * mg / (cos(θ) - μs sin(θ), resulting in a required force of approximately 210.224 N. Understanding the components of forces in both x and y directions is crucial for solving such physics problems.
xthursday
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This is a problem from my Physics 101 class that I'm having trouble solving. I'm not sure what equations I'm supposed to use.

Homework Statement


To move a large crate across a rough floor, you push on it with a force F at an angle of 21° below the horizontal. Find the force necessary to start the crate moving, given that the mass of the crate is m = 34 kg and the coefficient of static friction between the crate and the floor is 0.48.


Homework Equations


N= mgcosѲ = 311.386N
W= mgsinѲ = 119.53N
fs,max = mews*N = 149.465N
ΣF = N + fs + W


The Attempt at a Solution


Eek I don't know! That's why I'm posting!
 
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xthursday said:
This is a problem from my Physics 101 class that I'm having trouble solving. I'm not sure what equations I'm supposed to use.

Homework Statement


To move a large crate across a rough floor, you push on it with a force F at an angle of 21° below the horizontal. Find the force necessary to start the crate moving, given that the mass of the crate is m = 34 kg and the coefficient of static friction between the crate and the floor is 0.48.


Homework Equations


N= mgcosѲ = 311.386N
W= mgsinѲ = 119.53N
fs,max = mews*N = 149.465N
ΣF = N + fs + W


The Attempt at a Solution


Eek I don't know! That's why I'm posting!
Xthursday, welcome to the Forums! You have shown some attempt, but you must show more. This is a horizontal floor. The applied force (call it P) is at the angle noted. When summing forces acting on an object, you should look at the x and y components of the forces separately. In the x direction, you have the x component of the applied force acting right, and the friction force acting left. In the y direction, the weight acts straight down, the Normal force acts straight up,, and there is also the y component of the applied force acting down. Use Newton 1 in the y direction and in the x direction to solve for N and P. Please show your work.
 
With some help from other examples I figured out which equation I need to be using:
F= μs * mg / (cosѲ - μs sinѲ)
F= (.48)*(34kg)*(9.81m/s²) / [(cos(21) -.48(sin(21))]
F= 210.224 N
 

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