Frictional Force issue

  • #1

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?
 

Answers and Replies

  • #2
35,908
12,729
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.
 
  • #3
so f= 30 x 1.6=48
110-48=62N
correct ?

subtract because it moving in the negative direction, right?
 
  • #5
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?
 
  • #6
35,908
12,729
Which formulas do you have for (maximal) static friction?
 
  • #7
i do not, i was curious to find out which one to use?

do i have to find the net force?
 
  • #9
i have fs(max) = c0efficient of static friction x normal force x 9.81
 
  • #10
35,908
12,729
normal force x 9.81
That part is not correct, even if you add the missing units.
 

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