Calculate Max Speed of Truck on Unbanked Curve Without Eggs Sliding

[Doc Al]

wouldn't that be mg?

Sure. That's all you need.

 

[Paymemoney]

but how do i know the mass??

 

[ideasrule]

You don't need it. Use call the mass "m" and you'll see it cancels out.

 

[Paymemoney]

so if i use the F= N - mg

mg = N - mg

9.8 = N -9.8

N= 19.6N

 

[Doc Al]

but how do i know the mass??

You don't and it doesn't matter. Just use N = mg.

so if i use the F= N - mg

mg = N - mg

9.8 = N -9.8

N= 19.6N

No. The vertical forces add to zero, so 0 = N - mg, thus N = mg.

Now go back to analyzing the friction force. Now that you have an expression for N, how would you express the maximum static friction? Plug that into Newton's 2nd law.

 

[Paymemoney]

so if i use the F= N - mg

mg = N - mg

9.8 = N -9.8

N= 19.6N

however when i done it this way and i found the velocity to be the right answer as the book.

To answer your question would it be fsmax = \mu * mg

 

[Doc Al]

however when i done it this way and i found the velocity to be the right answer as the book.

You might get lucky now and then, but that method makes no sense. Learn how to do it right, then you're good to go no matter what.

To answer your question would it be fsmax = \mu * mg

OK. That's the net force providing the centripetal acceleration. Use Newton's 2nd law to find the velocity.

 

[Paymemoney]

do i use the equation F=\frac{mv^2}{r}??

 

[Doc Al]

do i use the equation F=\frac{mv^2}{r}??

Yes, where F is the friction force (in this case).

 

[Paymemoney]

ok i understand so:
\mu * N = \frac{mv^2}{r}

r(\mu * mg) = mv^2

cancel m's


r * \mu * g = v^2

then sub in values to get final answer of 14.3m/s

 

[Doc Al]

Exactly.