What is the Coefficient of Static Friction in This Physics Problem?

[HHenderson90]

Homework Statement

A child pulls a string attached to a block of mass m = 1.5 kg at an angle θ = 35.5° with an increasing amount of force. When she pulls with a force F = 7.39 N the block just begins to move. What is the coefficient of static friction between the block and floor?

Homework Equations

F_Static=U_static*F_normal

The Attempt at a Solution

I first simply called the force she pulls with F_Static, and divided it by the Normal Force (M*G)
This was wrong.
I then thought that maybe only the force in the x-direction would be considered the static force, I found that by FCos(theta). I then divided this by the force normal, and again no luck.
I then thought the force applied by the child in the y direction might be subtracted from the force normal? This didn't work either.
What do I do?

 

[Dick]

Your second and third approaches combined sound right. The force in the y direction should subtract from the normal force and the force in x direction is the force contributing to making the block slide. It would be a good idea to show your work and the numbers you got.

 

[HHenderson90]

I ended up getting it correct by combing my second and third approach. Found the x-component of the force applied by 7.39Cos(35.50 which was 6.0163N and I called this F_Static. I found the F_N by M*A which was 15*9.81=14.715
Subtracting the y-component of the applied force (7.39sin(35.5)) I got the F_N as 10.4336.
F_Static= U_s*F_N
Dividing F_Static by F_N I got the answer of .577.

There is a second part to the problem that I am having trouble with now.
1. Homework Statement
Once the block starts to move, it accelerates at a rate of 1.13 m/s2. What is the coefficient of kinetic friction between the block and floor? 2. Homework Equations
Fnet=M*A
F_Kinetic=Uk*Fn
3. The Attempt at a Solution

I think my issue here is what constitutes the Fnet.
I found Fnet by multiplying M*A 1.5*1.13=1.695N
I called this the force applied, considering that the x component would be equal to the force of static friction.
Finding the x component- 1.695cos(35.5)=1.38N
I then took the Fn from before, and subtracted the y component from this force
14.715-1.695sin(35.5)=13.73N
I then used the Fk=UkFn
1.38=Uk13.73=.100

This answer is wrong.

 

[Dick]

I ended up getting it correct by combing my second and third approach. Found the x-component of the force applied by 7.39Cos(35.50 which was 6.0163N and I called this F_Static. I found the F_N by M*A which was 15*9.81=14.715
Subtracting the y-component of the applied force (7.39sin(35.5)) I got the F_N as 10.4336.
F_Static= U_s*F_N
Dividing F_Static by F_N I got the answer of .577.

There is a second part to the problem that I am having trouble with now.
1. Homework Statement
Once the block starts to move, it accelerates at a rate of 1.13 m/s2. What is the coefficient of kinetic friction between the block and floor?


2. Homework Equations
Fnet=M*A
F_Kinetic=Uk*Fn



3. The Attempt at a Solution

I think my issue here is what constitutes the Fnet.
I found Fnet by multiplying M*A 1.5*1.13=1.695N
I called this the force applied, considering that the x component would be equal to the force of static friction.
Finding the x component- 1.695cos(35.5)=1.38N
I then took the Fn from before, and subtracted the y component from this force
14.715-1.695sin(35.5)=13.73N
I then used the Fk=UkFn
1.38=Uk13.73=.100

This answer is wrong.

Your applied x and y forces and normal force are exactly the same as in the first problem. What's different is that the horizontal force Fx and the force of friction Fk combine to accelerate the block. So (Fx-Fk)=m*a. Use that to find Fk.

 

[Nickie]

As a scientist, it is important to approach problems like this with a systematic and logical approach. In this case, we can use the equations and principles of static friction to find the coefficient of friction between the block and the floor.

First, we can start by drawing a free body diagram of the block, including all the forces acting on it. In this case, we have the normal force, the force applied by the child, and the force of static friction. We can also label the components of each force, using the given angle and force values.

Next, we can apply Newton's Second Law, which tells us that the sum of all forces acting on the block must equal its mass times its acceleration (which is zero in this case since the block is not moving). This will give us a mathematical equation to solve for the coefficient of static friction.

We can then use the equation for static friction, F_Static = μ_Static * F_Normal, and substitute in the values we have from the free body diagram and the given information. This equation relates the force of static friction to the coefficient of static friction and the normal force between the block and the floor.

From here, we can solve for the coefficient of static friction by rearranging the equation and plugging in the values we have. This will give us a numerical value for the coefficient of static friction between the block and the floor.

It is important to note that the normal force is equal to the weight of the block, which is given by the mass and the acceleration due to gravity. This means that the normal force will not change as the force applied by the child increases. Only the force of static friction will change, until it reaches the maximum value and the block begins to move.

In conclusion, by using the principles of static friction and Newton's Second Law, we can systematically solve for the coefficient of static friction between the block and the floor. This approach allows us to accurately and logically solve problems in physics and other scientific fields.