Physics Problem: Solving Friction Coefficient on Lawnmower

[Inspector Gadget]

I have a test in my physics class early this week, and there are two problems on the review sheet I am having problems getting...both of them having a frictional coefficient...

Here is the first one that I think I understand but the answer I'm getting doesn't exactly make sense...

PROBLEM: An 80N force is applied to the handle of a lawnmower. The handle makes an angle of 40º with the ground. If the coefficient of friction between the lawnmower and the ground is 0.15, find the acceleration of the 30kg lawnmower.

I've attached an image below and I'll explain what I have on it...everything in black is given information and what's in red I figured out, which I'll explain.

Using the SOH-CAH-TOA functions, I found that the vertical force was 61.3 N, and the horizontal force is 51.4 N, and our teacher has as round all forces in Newtons to the nearest whole number, so that's 61 N and 51 N.

Next, the downward force of the lawnmower is...
W=m*g=(30 kg)(9.8 m/sec²)=294 N

The normal / perpindicular force pushing up on the lawnmower has to be equal, so you have the 294N going down from the weight of the lawnmower, plus the 61N of downward force from pushing the lawn mower, that's 294N + 61N = 355N.

frictional force = (frictional coefficient)(normal force)=(.15)(355)=53N

There's a 51N force that we found from pushing the lawnmower pushing forward. But from what I found about, you need 53N to overcome friction, so I'm finding the lawn mower isn't even moving. Maybe what I did is right and it's true, but I'd like someone to just look over what I did and either correct or confirm.

Thanks in advance.

 

[RichD]

Your answer looks right to me. I'd say it wasn't going to accelerate from rest in the curretn configuration too.

 

[No Name Required]

Check your resolving of the 81N force into its horizontal and vertical components, that's where ure mistake is.

I get the acceleration as approximately 0.31 m/s/s