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heyyalliee · Sep 30, 2010, 11:21 AM

To move a large crate across a rough floor, you push on it with a force F at an angle of 21 degrees below the horizontal.

Q.Find the force necessary to start the crate moving, given that the mass of the crate is 32 kg and the coefficient of static friction between the crate and the floor is 0.59.

I thought I could solve for N by N=mgcos(theta) but I think I have it all wrong.
Help?

Unknown008 · Sep 30, 2010, 11:40 AM

If I understood well, the crate is on horizontal ground.

Find the maximum frictional force on the crate.

$F =\mu R = 0.59 \times 32g$

Now the force should exceed this, but since it's applied at an angle, you have to break it into components.

Horizontal component = $Fcos(21)$

Then, a vertical component $=Fsin(21)$

Now, since you have an additional vertical component, the reaction of the crate on the ground is decreased.

$R - Fsin(21)= mg$

So, we get:

$F =\mu R = 0.59 \times (mg + Fsin(21))$

Hence:

$Fcos(21)=0.59 \times (mg + Fsin(21))$

heyyalliee · Oct 1, 2010, 07:08 AM

Okay I follow you just want to make sure I'm doing this right.

My horizontal component was 17.63 and Vertical 6.766.

To solve for mg I did 32g+ 6.766 and got 38.77.

Now to find the force necessary to move crate
do I do this? ( Assuming that the x in the problem you gave me was multiplication)
.59(38.77-6.766)
I get 18.88.

Unknown008 · Oct 1, 2010, 10:15 AM

I'll try re-explain. Maybe I took the wrong approach earlier.

As you can see you don't know the value of the force on the left nor the value of the frictional force.

On the left (in blue) is the horizontal component of the force applied and on the floor on the right is the frictional force.

At equilibrium, the blue line and the frictional force are the same.

$F cos(21) = \mu R$

Okay, now, we need R.

We can find R by resolving the vertical components. (I didn't show them in the picture)

Downwards, you get mg + Fsin(21)

Upwards, you get R.

So, we get: $R = mg + Fsin(21)$

Use that in the horizontal forces equation:

$F cos(21) = \mu (mg + Fsin(21))$

From this, you will find the value of F. (I'm taking g = 10 m/s^2)

$Fcos(21) = 0.59 ((32\times10) + Fsin(21))$

$Fcos(21) = 188.8 + 0.59Fsin(21))$

$Fcos(21) - 0.59Fsin(21) = 188.8$

$F(cos(21) - 0.59sin(21)) = 188.8$

$F = \frac{188.8}{cos(21) - 0.59sin(21)}$

$F =261.4 N$

So, as soon as the force applied exceeds 261.4 N, the crate will start moving.

heyyalliee · Oct 3, 2010, 12:54 PM

Sorry I was confused for sure. I try to oversimplify too much. It's the components that always confuse me. I have a hard time visualizing it and then finding the right equations. I appreciate your help!