- #1

- 5

- 0

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

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- Thread starter chimbooze
- Start date

- #1

- 5

- 0

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

- #2

Doc Al

Mentor

- 45,250

- 1,601

No, both sled and box are accelerating. (They're being pulled by a rope.)a of box is zero since it is not moving

Hint: What is the maximum static friction between box and sled?

- #3

- 5

- 0

Ff = mu x n

Ff = .5(9.8)(5)

But that isn't the right answer.

- #4

Doc Al

Mentor

- 45,250

- 1,601

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?But that isn't the right answer.

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

- #5

- 5

- 0

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.

- #6

Doc Al

Mentor

- 45,250

- 1,601

Excellent!

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