Tension Force problem involving friction

AI Thread Summary
A horizontal rope pulls a 20 kg sled with a 5 kg box on it across frictionless snow, and the goal is to determine the maximum tension force before the box slips. The coefficient of static friction between the box and sled is 0.5, leading to a maximum static friction force calculated using Ff = mu x n. The box does not accelerate, so the tension must equal the maximum friction force to prevent slipping. After applying Newton's second law, the maximum acceleration of the sled is found to be 4.9 m/s², resulting in a tension force of 73.5 N. The calculations confirm the correct understanding of the forces involved.
chimbooze
Messages
5
Reaction score
0
1. A horizontal rope pulls a 20kg wood sled across frictionless snow. A 5 kg wood box rides on the sled. The coefficient of static friction for wood on wood is 0.5. What is the largest tension force for which the box doesn't slip?

m = 5 kg box, 20 kg sled
mu = .5
T = ?

2.
F = ma
Ff= mu x n

3.
F = ma
a of box is zero since it is not moving

T - mu x (20 + 5) x (9.8) = 0
 
Physics news on Phys.org
chimbooze said:
a of box is zero since it is not moving
No, both sled and box are accelerating. (They're being pulled by a rope.)

Hint: What is the maximum static friction between box and sled?
 
I partially get it. If Tension force is greater than the static friction force, then the box will move. So just find friction force.

Ff = mu x n
Ff = .5(9.8)(5)

But that isn't the right answer.
 
chimbooze said:
But that isn't the right answer.
That's just one step toward the answer. Now that you know the maximum friction force on the box, what is the maximum acceleration that the sled can have without the box slipping?

To get the final answer, you'll need to apply Newton's 2nd law twice.
 
So:

sum of F = ma
Ff = ma
.5(9.8)(5) = 5(a)

a = 4.9

Tension of rope: T = ma

T = (10 + 5)(4.9)
T = 73.5

Thanks a lot. I got it correct.
 
Excellent!
 
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