# Frictional Force issue

pennywise1234

## Homework Statement

You pull on a 30 kg box with a horizontal force of 110 N to the right. (i) If there are no other forces acting on this box, what is the resulting acceleration? (ii) If the resulting acceleration of the block is only 1.6 m/s, what is the magnitude and direction of the frictional force acting on the block

## The Attempt at a Solution

For (i) a- f/m and I got a=3.67

i have no idea how to solve for (ii) i thought it would f=m times a, but i have no idea where to start/solve any help, please?

Mentor
i thought it would f=m times a
You know m and a, so you can calculate F, the net force acting on the block. This net force has two components.

pennywise1234
so f= 30 x 1.6=48
110-48=62N
correct ?

subtract because it moving in the negative direction, right?

Mentor
Correct.

pennywise1234
pennywise1234
im also stuck on static frictional force question

The static co-efficient of friction between a football blocking sled (90 kg) and grass is 0.56. (i) If an80 kg coach stands on top of the blocking sled, what is the maximal static frictional force?

any hints on the formula to apply?

Mentor
Which formulas do you have for (maximal) static friction?

pennywise1234
i do not, i was curious to find out which one to use?

do i have to find the net force?

Mentor
i do not, i was curious to find out which one to use?
There should be exactly one.
do i have to find the net force?
You have to find the maximal force before block+trainer start to move.

pennywise1234
i have fs(max) = c0efficient of static friction x normal force x 9.81

Mentor
normal force x 9.81
That part is not correct, even if you add the missing units.