How Do You Calculate the Force Needed to Move a Block Across the Ground?

[monkeyhead]

Homework Statement

Hi,
The question I am a bit confused about is this:

A block is dragged across the ground by a light cable,attached to a winch. Th block is initally at rest.

Information given:
Volume: 6m³
Density 2875 kg/m³
Coefficent of friction between block and the ground: 0.4
Diameter: 1.4m

Find the force that must be applied to the block to overcome friction and cause it to accelerate at a rate of 0.4m/s²

The Attempt at a Solution

So I've first got to find the mass right? m=D.V= 2875*6= 17,250 KG

Then I can find its weight. w=m.g = 17,250*9.81= 169,222.5 N

Now I need to find the force acting down on the block right? Fa= m.a
=17,250*0.4= 6900N

And the frictional force acting Frictionalforce= μ*w = 0.4*169,222.5= 67,689N

Then the total force should equal: Tforce=√Frictional force²+Fa²= √67689²+6900²= 68,039.77308 right?

Have I gone wrong somewhere?

Many thanks
Matt

 

[mbrmbrg]

You're doing OK until you add forces.
Newton's seond law states:
\Sigma F = ma
Sigma means add up the forces algebraicly. Squaring stuff is when you have forces acting along x- and y- axes.

 

[monkeyhead]

So F - friction force = ma
where F=? Frictional force=67689 ma= (17250*0.4)=6900

So F - 67689 = 6900?

Many thanks
Matt

 

[mbrmbrg]

So F - friction force = ma

Yes.

where F=? Frictional force=67689 ma= (17250*0.4)=6900

So F - 67689 = 6900?

Don't really understand what you have written here. But if you solve your last line for F, you and I will get very similar answers.

 

[monkeyhead]

Would that mean F= 6900+67689=74589? Right?
Many thanks
Matt

 

[mbrmbrg]

The number looks good to me--just slap on units and a direction, and check significant figures, and that's it.

 

[monkeyhead]

Thanks!
Out of interest, how would I go about finding out the angle?
Many thanks
Matt

 

[mbrmbrg]

Everything done solved for the x- component of the force. No information was given about the y-component.