A steel shuttle is pulled along a rail, what is its initial acceleration?

Click For Summary
The initial acceleration of a steel shuttle pulled by a 25N tension force at a 45-degree angle is calculated using F/m = a, resulting in 32.89 m/s². However, the user encountered issues when attempting to find the x component of acceleration, leading to confusion about the role of friction. It was identified that friction must be considered, with a coefficient of 0.8 for steel on steel. The corrected acceleration formula incorporates this friction, adjusting the calculation accordingly. Understanding the impact of friction is crucial for accurate results in this scenario.
potatotimer
Messages
2
Reaction score
0
Homework Statement
Question: a machine has a 760g steel shuttle that is pulled along a square steel rail by an elastic cord. the shuttle is released when the cord has a 25N tension at a 45-degree angle what is initial acceleration?
Relevant Equations
F = ma or F/m = a
it sounds like you just need to use F/m = a which gives 25 / .76 = 32.89 = a
this didn't work so I tried to get the x component of acceleration so I did 32.89cos(45) = 23.26 and this didn't work.

what am I missing, the force is tension right so it should be 25N?
 
Physics news on Phys.org
potatotimer said:
Homework Statement:: Question: a machine has a 760g steel shuttle that is pulled along a square steel rail by an elastic cord. the shuttle is released when the cord has a 25N tension at a 45-degree angle what is initial acceleration?
Relevant Equations:: F = ma or F/m = a

it sounds like you just need to use F/m = a which gives 25 / .76 = 32.89 = a
this didn't work so I tried to get the x component of acceleration so I did 32.89cos(45) = 23.26 and this didn't work.

what am I missing, the force is tension right so it should be 25N?
Two possibilities come to mind. First, have you rounded to an appropriate number of significant figures? Second, have you been studying friction and, if so, would the friction of steel on steel come into play?
 
jbriggs444 said:
Two possibilities come to mind. First, have you rounded to an appropriate number of significant figures? Second, have you been studying friction and, if so, would the friction of steel on steel come into play?
yeah its friction
rereading made me realize that friction does need to be considered which for a steel on steel collision u = 0.8
so the acceleration is 25cos(45) - .8*.76*9.8
thanks for the help :)
 
potatotimer said:
- .8*.76*9.8
Two errors in there.
 
Thread 'Chain falling out of a horizontal tube onto a table'
My attempt: Initial total M.E = PE of hanging part + PE of part of chain in the tube. I've considered the table as to be at zero of PE. PE of hanging part = ##\frac{1}{2} \frac{m}{l}gh^{2}##. PE of part in the tube = ##\frac{m}{l}(l - h)gh##. Final ME = ##\frac{1}{2}\frac{m}{l}gh^{2}## + ##\frac{1}{2}\frac{m}{l}hv^{2}##. Since Initial ME = Final ME. Therefore, ##\frac{1}{2}\frac{m}{l}hv^{2}## = ##\frac{m}{l}(l-h)gh##. Solving this gives: ## v = \sqrt{2g(l-h)}##. But the answer in the book...

Similar threads

  • · Replies 4 ·
Replies
4
Views
2K
Replies
2
Views
2K
Replies
1
Views
1K
  • · Replies 38 ·
2
Replies
38
Views
4K
Replies
2
Views
1K
  • · Replies 5 ·
Replies
5
Views
2K
  • · Replies 42 ·
2
Replies
42
Views
5K
Replies
15
Views
4K
  • · Replies 3 ·
Replies
3
Views
2K
  • · Replies 56 ·
2
Replies
56
Views
4K