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

AI Thread 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 'Variable mass system : water sprayed into a moving container'
Starting with the mass considerations #m(t)# is mass of water #M_{c}# mass of container and #M(t)# mass of total system $$M(t) = M_{C} + m(t)$$ $$\Rightarrow \frac{dM(t)}{dt} = \frac{dm(t)}{dt}$$ $$P_i = Mv + u \, dm$$ $$P_f = (M + dm)(v + dv)$$ $$\Delta P = M \, dv + (v - u) \, dm$$ $$F = \frac{dP}{dt} = M \frac{dv}{dt} + (v - u) \frac{dm}{dt}$$ $$F = u \frac{dm}{dt} = \rho A u^2$$ from conservation of momentum , the cannon recoils with the same force which it applies. $$\quad \frac{dm}{dt}...
Back
Top