Mass of a box been pushed taking friction into account.

CBM

May 2019
4
0
This feels like a really basic question so i feel like I'm making an assumption that I shouldn't be.

Max fs=us*N
Fk=uk*N
N=mg

For the box to start to move Fk must equal Fs. So I can rearrange for m from there us*75*9.81=uk*9.81*m but this answer isn't ending up anywhere near the correct answer

Idk how to delete this.
There was no problem with it I just can't use a calculator right apparently.
 

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Last edited:
Apr 2017
518
125
no need for formula , no need for g ...

The 'grip' per kilo of the 75kg person (0.8) is 4 times the 'grip' per kilo of the box (0.2) so the box can be 4 times more massive than the person.
 
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Oct 2017
536
255
Glasgow
The snippet you uploaded doesn't actually describe the entire question, just the context. What does the question ask you to find out?
 
Oct 2017
536
255
Glasgow
Everything you need is there benit .. they ask for the maximum mass of the box.
Okay, thanks. I was a bit puzzled because it just seems like it's going to state what it is and then doesn't, but maybe the question is phrased as one of those "the answer is this... now prove it" ones and the OP wanted to keep the statement hidden.

Your answer is fine, but here's the formulae anyway in case the OP is interested.

Assuming that the pushing force that the person exerts on the box is the same as the frictional force from the person's shoes (Newton's third law), the maximum force the person can apply is:

\(\displaystyle F_{person} = \mu_{person} m_{person} g\)

The box will move when this force exceeds the friction force of the box on the floor.

\(\displaystyle F_{box} = \mu_{box} m_{box} g\)

The maximum mass of the box is therefore the mass that makes the box's friction force equal to the person's pushing force, i.e.

\(\displaystyle F_{person} = F_{box}\)

Substitute and rearrange for \(\displaystyle m_{box}\):

\(\displaystyle m_{box} = m_{person} \cdot \frac{\mu_{person}}{\mu_{box}} = 75 \times \frac{0.8}{0.2} = 300 kg\)
 
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