# Finding Acceleration of a cart. NEW

## Homework Statement

A grocery cart is being pushed with a force of 450 N at an angle of 30 degrees to the horizontal. If the mass of the cart and the groceries is 42 kg:
a) Find force of friction if coefficient of friction is 0.6
b) Find acceleration of cart.

Ff=uFnormal
F=ma
Fn=mg

## The Attempt at a Solution

I figured out part A and I got 3.8 X 10^2 N as force of friction.

For part B, I tried to find force applied.
Fa = 450 N X cos30 degrees
= 389.7 N

and then find net force:

Fnet = 389.7 - (3.8 X 10^2)
= 9.7

and then find acceleration:

a = 9.7 / 42
= 0.23

But the answer is wrong and I checked from my book. Can someone tell me if I am right on part B of this question. Thanks.

it is so much easier if you post the math that got you to the answer,
assuming the force is downward on the cart,
N=mg+f*sin 30 so N=.5*450+mg=9.8*42+225=636.6
can you do it from here?

But I don't understand what to do with 636.6. I actually did that before but then I thought that's wrong since the number is too big.

I am guessing that probably I subtract 450 from 636 to get net force. But I am too confused. Can you please help more? Thanks.

sure. but lets make certain you understand the first part:

friction resists motion, but it only dependent on the force pushing the objects into contact. that force is the Normal or perpendicular force. So if we have a weight of 10okg*g, the normal force directed straight up is 980N. When we multiply this by the coefficient of friction, it is now a horizontal resistance of magnitude say .6*980.

Last edited:
hi. i'm currently doing this question too.
but for some reason i am not getting part A. could you explain what you did to get 3.8 x 10^2 N?
what i did was:
Fn=mg
Fn=(42kg) (9.81m/s^2)
Fn=412.02N
Ff=0.6(412.02N)
Ff=247.212N

my answer is no near to the book's answer. what am i doing wrong?!!