Solving Mechanics Friction Problem: How to Calculate Contact Force

In summary, the conversation revolves around finding the total contact force and coefficient of friction for a system in static equilibrium on an incline. The correct values for F and R are 33.7N and 85.8N, respectively, and the coefficient of friction is 0.393. The person seeking help is advised to consult their instructor for further assistance.
  • #1
Shah 72
MHB
274
0
View attachment 11145
It's the 10th question. Iam getting the ans for R=90cos34= 74.6 and F=-30.3N, so I get total contact force by using the formula c= sqroot(f^2+R^2)= 80.52N
The ans in the textbook is 92.2N. Iam not getting the right ans for coefficient of friction also which is 0.393
 
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  • #2
Attachment doesn't show ...
 
  • #3
skeeter said:
Attachment doesn't show ...
Iam so sorry.
20210513_222544.jpg
 
  • #4
I get $\mu = 0.393$ ...

FBD_incline.jpg
 
  • #5
  • #6
Shah 72 said:
Can you pls show me the working

no.

I've provided the correct FBD ... this one is on you.

Hint: the system is in static equilibrium in both the parallel & perpendicular directions respective to the incline.
 
  • #7
skeeter said:
no.

I've provided the correct FBD ... this one is on you.

Hint: the system is in static equilibrium in both the parallel & perpendicular directions respective to the incline.
R=90cos34
F=90sin34-20
Mu=0.406
I don't get the right ans
 
  • #8
R is incorrect … what about the component of the 20N applied force perpendicular to the incline?

F net is incorrect … the component of the 20N applied force parallel to the incline is not 20.
 
  • #9
skeeter said:
R is incorrect … what about the component of the 20N applied force perpendicular to the incline?

F net is incorrect … the component of the 20N applied force parallel to the incline is not 20.
R+20sin34= 90cos34
Is this the way to do?
 
  • #10
skeeter said:
R is incorrect … what about the component of the 20N applied force perpendicular to the incline?

F net is incorrect … the component of the 20N applied force parallel to the incline is not 20.
20210514_092813.jpg
 
  • #11
The problem statement says the 20N force is horizontal … that would be w/respect to the Earth’s surface, not the surface of the incline.

The reaction force in your diagram is also incorrect.

Your FBD looks nothing like the one I posted.

If you still can’t get the correct value for $\mu$, I recommend you seek assistance from your instructor.

I’m done.
 
  • #12
skeeter said:
The problem statement says the 20N force is horizontal … that would be w/respect to the Earth’s surface, not the surface of the incline.

The reaction force in your diagram is also incorrect.

Your FBD looks nothing like the one I posted.

If you still can’t get the correct value for $\mu$, I recommend you seek assistance from your instructor.

I’m done.
Ok no problem. I will try to work it out. Thank you!
 
  • #13
skeeter said:
The problem statement says the 20N force is horizontal … that would be w/respect to the Earth’s surface, not the surface of the incline.

The reaction force in your diagram is also incorrect.

Your FBD looks nothing like the one I posted.

If you still can’t get the correct value for $\mu$, I recommend you seek assistance from your instructor.

I’m done.
I got the ans. So F= 33.7N and R=85.8N
I got the ans 0.393.
Thank you so much!
 

Related to Solving Mechanics Friction Problem: How to Calculate Contact Force

1. What is friction and why is it important to calculate contact force?

Friction is a force that opposes the motion of an object when it comes into contact with another object. It is important to calculate contact force because it can affect the movement and stability of objects, and understanding it is crucial in designing and building structures and machines.

2. How do you calculate contact force in a mechanics friction problem?

To calculate contact force, you need to know the coefficient of friction, the normal force, and the angle of the surface. The formula for contact force is: F = μNcosθ, where F is the contact force, μ is the coefficient of friction, N is the normal force, and θ is the angle of the surface.

3. What is the difference between static and kinetic friction?

Static friction is the force that keeps an object at rest when a force is applied to it, while kinetic friction is the force that opposes the motion of an object that is already in motion. Static friction is typically greater than kinetic friction.

4. How does the coefficient of friction affect the calculation of contact force?

The coefficient of friction is a measure of how much friction exists between two surfaces. A higher coefficient of friction means there is more resistance to motion, resulting in a higher contact force. Conversely, a lower coefficient of friction means there is less resistance to motion, resulting in a lower contact force.

5. Can you provide an example of a mechanics friction problem and how to solve it?

Sure, let's say you have a box weighing 20 kg on a flat surface with a coefficient of friction of 0.5. The normal force on the box is equal to its weight, which is 20 kg x 9.8 m/s^2 = 196 N. The contact force can be calculated using the formula F = μNcosθ, where θ = 0 degrees for a flat surface. Therefore, the contact force is F = 0.5 x 196 N x cos 0 = 98 N. This means that there is a contact force of 98 N opposing the motion of the box.

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