# 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.

Related Introductory Physics Homework Help News on Phys.org
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?!!