Max Tension Force on Horizontal Rope Pulling 10kg Wood Sled

[Comtrend]

I have a question where a horizontal rope pulls on a 10kg wood sled over a frictionless surface. On that sled is a wood box. It is asking what the largest tension force in the rope would be where the box doesn't slip off.

Air resistance/Drag is neglected. Wouldn't the box remain on the sled indefinetely as long as it didn't accelerate to fast? If not then I would use the coefficient of static friction right?

 

[Andrew Mason]

I have a question where a horizontal rope pulls on a 10kg wood sled over a frictionless surface. On that sled is a wood box. It is asking what the largest tesion force in the rope would be where the box doesn't slip off.

Air resistance/Drag is neglected. Wouldn't the box remain on the sled indefinetely as long as it didn't accelerate to fast? If not then I would use the coefficient of static friction right?

You would need to know the coefficient of static friction between the box and sled to find the maximum static friction force. That is the maximum force that the sled can apply to the box. If the tension exceeds that force, the box will slide off.

AM

 

[clive]

Andrew is right, you need the frictional force between the sled and the box.

for the box:
F_f=m_{box} a

and for the sled
F-F_f=m_{sled} a

As we can see, we have 3 unknows and only 2 eqs. So you must provide some information about the friction at the contact surface.

 

[Comtrend]

I got 14.72N with a coefficient of static friction of 0.3. So is the Ff equal to the force of Tension because the sled is on a frictionless surface?

So then Tension equals 14.72N?

 

[Comtrend]

For force of friction for the box I used F_f=mN


I don't know how to get mew so it is just an m.

 

[clive]

The tension you need is F (in my eqs). F_f (the frictional force) and F (the tension in the rope) are different.

You have to find F from the eq. below:
\frac{F-F_f}{F_f}=\frac{m_{sled}}{m_{box}}
and you'll obtain the tension in the rope.