Finding Force via Static Friction

In summary, to move a large crate across a rough floor at an angle of 21° below the horizontal, with a mass of 34 kg and a coefficient of static friction of 0.48, a force of 210.224 N is necessary to start the crate moving. This can be calculated using the equation F= μs * mg / (cosѲ - μs sinѲ).
  • #1
xthursday
4
0
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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  • #2
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.
 
  • #3
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
 

1. What is static friction?

Static friction is a type of force that occurs when two objects are in contact with each other and one object is not moving relative to the other. It acts in the opposite direction of the force trying to move the object and prevents it from moving.

2. How is static friction different from kinetic friction?

Static friction is the force that prevents an object from moving, while kinetic friction is the force that opposes the motion of an object that is already moving. Static friction is typically greater than kinetic friction.

3. How do you calculate the force of static friction?

The force of static friction can be calculated using the equation F = μsN, where μs is the coefficient of static friction and N is the normal force between the two objects in contact.

4. What factors affect the force of static friction?

The force of static friction is affected by the coefficient of static friction, the normal force, and the roughness of the surfaces in contact. The force of static friction also increases as the applied force trying to move the object increases.

5. How can static friction be useful in everyday life?

Static friction is useful in everyday life in many ways, such as allowing us to walk without slipping, keeping objects in place on inclined surfaces, and allowing cars to grip the road while driving. It also helps in activities like playing sports, writing, and using tools.

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