The maximum speed the truck can go without sliding

AI Thread Summary
The discussion revolves around calculating the maximum speed a truck can achieve without a crate of eggs sliding off while rounding a circular bend with a radius of 35 meters. The coefficient of static friction between the crate and the truck is given as 0.6. The calculations involve using the formula for static friction and centripetal force, leading to the equation v=√(μsgr). After substituting the values, the resulting maximum speed is calculated to be 14.35 m/s. A minor correction was noted regarding the terminology used in the explanation.
goli12
Messages
11
Reaction score
0

Homework Statement


A crate of eggs is located on the back of a truck. The truck rounds a circular bend in the road with radius of 35 meters. If the coefficient of static friction between the crate and the truck is 0.6, what is the maximum speed the truck can go without the crate sliding?

Homework Equations


F=mv2/r
Fs=μsn

The Attempt at a Solution


Hey guys, can you please check my answer for this question, I'm not entirely sure it is correct...

Firstly, I took the normal force of the truck to get n=mg, as it is not moving vertically. This results in the force of static friction (Fs) to be: Fs=μsmg.

Secondly, this static friction is providing the centripetal force of the truck to go around in a circular motion and so resulting in the following equation: Fs=F=mv2/r = μsmg.

So with algebra I made velocity the subject. The masses cancel out leaving: v2/r = μsg and so resulting in v=√(μsgr)

Now I enter the variables v=√(0.6*9.8*35), giving the velocity of 14.35ms-1.
 
Physics news on Phys.org
Looks fine to me :)
 
Solution looks ok, however, substitute the word 'truck' with 'crate' in your attempt.
 
PhanthomJay said:
Solution looks ok, however, substitute the word 'truck' with 'crate' in your attempt.

Ahh yes, my mistake many thanks :-p
 
I multiplied the values first without the error limit. Got 19.38. rounded it off to 2 significant figures since the given data has 2 significant figures. So = 19. For error I used the above formula. It comes out about 1.48. Now my question is. Should I write the answer as 19±1.5 (rounding 1.48 to 2 significant figures) OR should I write it as 19±1. So in short, should the error have same number of significant figures as the mean value or should it have the same number of decimal places as...
Thread 'A cylinder connected to a hanging mass'
Let's declare that for the cylinder, mass = M = 10 kg Radius = R = 4 m For the wall and the floor, Friction coeff = ##\mu## = 0.5 For the hanging mass, mass = m = 11 kg First, we divide the force according to their respective plane (x and y thing, correct me if I'm wrong) and according to which, cylinder or the hanging mass, they're working on. Force on the hanging mass $$mg - T = ma$$ Force(Cylinder) on y $$N_f + f_w - Mg = 0$$ Force(Cylinder) on x $$T + f_f - N_w = Ma$$ There's also...
Back
Top