Please help me with this physics problem on static friction

AI Thread Summary
To solve the physics problem regarding the coefficient of static friction between a crate and a truck bed, start by drawing a free body diagram (FBD) to identify all forces acting on the crate. Use Newton's laws to establish equations for both horizontal and vertical force components, considering the crate's weight and normal force. Calculate the acceleration of the truck using the formula Vf^2 - Vi^2 = 2ad, assuming the truck stops after 28.7 m. The relationship between the force of friction and the normal force can be expressed as Force of friction = (normal force)(static coefficient). The solution ultimately leads to the correct coefficient of static friction, which is confirmed as 0.400.
redhawks
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I am pretty sure the answer might be A. Can someone help by showing the work needed to solve the problem. It would be greatly appreciated

Homework Statement



A crate is carried in a pickup truck traveling horizontally at 15.0 m/s. The truck applies the brakes for distance at 28.7 m while stopping with uniform acceleration. What is the coefficient of static friction between the crate and the truck bed, if the crate does not slide?

Homework Equations



A. 0.400
B. 0.365
C. 0.892
D. 0.656

The Attempt at a Solution



I'm not really sure how to start
 
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Hi redhawks,

redhawks said:
I am pretty sure the answer might be A. Can someone help by showing the work needed to solve the problem. It would be greatly appreciated

Homework Statement



A crate is carried in a pickup truck traveling horizontally at 15.0 m/s. The truck applies the brakes for distance at 28.7 m while stopping with uniform acceleration. What is the coefficient of static friction between the crate and the truck bed, if the crate does not slide?

Homework Equations



A. 0.400
B. 0.365
C. 0.892
D. 0.656

The Attempt at a Solution



I'm not really sure how to start


Draw a force diagram for the crate by itself, showing all forces that act on it directly and its acceleration. Then use Newton's laws to get two equations, one for the horizontal force components and one for the vertical force components.
 
Thanks, but i know to do the FBD. I think the distance part throws me off. If someone can go through the steps, it would really help me out.
 
i arrived at solution a also... although i can't give you a written out explanation, i can try to point you in the write direction...

a good first step is to draw out that FBD... for y components you should see you have the weight of the box (downward) and the normal force on the bottom of the box (upward). the x components should show the force of the box due to the acceleration (to the right), and the force friction countering the force of the box (to the left)

you need to figure out the force of the box due to accelartion
use
F=ma, and i used Vf^2-Vi^2=2ad to determine the acceleration a (here i will give you a little tip, i assumed the truck came to a stop after the 28.7m, so that Vf=0) This part of the problem is dynamic when looking at the problem in reference to the truck in motion on the road...

if you then focus your attention to just the box and the bed of the truck, there is no relative motion between the two and you can assume static equations...

here is used sum of x components F=0, i also used (Force of friction)=(normal force)(static coefficient)



let me know how it goes from here, if you need more help i will be glad to, well..., help hahahah
 
Thanks a lot. I GOT IT!:smile:
 
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Kindly see the attached pdf. My attempt to solve it, is in it. I'm wondering if my solution is right. My idea is this: At any point of time, the ball may be assumed to be at an incline which is at an angle of θ(kindly see both the pics in the pdf file). The value of θ will continuously change and so will the value of friction. I'm not able to figure out, why my solution is wrong, if it is wrong .
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