Magnitude and Direction of Frictional Force on a Moving Object

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Homework Help Overview

The discussion revolves around determining the magnitude and direction of the frictional force acting on a box being pulled with a specified force, as well as exploring a related problem involving static friction on a sled. The subject area includes dynamics and frictional forces.

Discussion Character

  • Mixed

Approaches and Questions Raised

  • Participants discuss calculating acceleration and net force, with some attempting to derive the frictional force from given parameters. Questions arise regarding the correct application of formulas for static friction and the relationship between forces.

Discussion Status

Some participants have provided calculations and attempted to clarify the relationships between forces, while others are seeking guidance on the appropriate formulas for static friction. There is an ongoing exploration of different interpretations and approaches to the problems presented.

Contextual Notes

Participants express uncertainty about the necessary formulas and the implications of static versus kinetic friction, indicating a need for further clarification on these concepts.

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

subtract because it moving in the negative direction, right?
 
Correct.
 
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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?
 
Which formulas do you have for (maximal) static friction?
 
i do not, i was curious to find out which one to use?

do i have to find the net force?
 
pennywise1234 said:
i do not, i was curious to find out which one to use?
There should be exactly one.
pennywise1234 said:
do i have to find the net force?
You have to find the maximal force before block+trainer start to move.
 
i have fs(max) = c0efficient of static friction x normal force x 9.81
 
  • #10
pennywise1234 said:
normal force x 9.81
That part is not correct, even if you add the missing units.
 

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