Question

You see the boy next door trying to push a crate down the sidewalk. He can barely keep it moving, and his feet occasionally slip. You start to wonder how heavy the crate is. You call to ask the boy his mass, and he replies " 48 kg ." From your recent physics class you estimate that the static and kinetic coefficients of friction are 0.90 and 0.40 for the boy's shoes, and 0.50 and 0.20 for the crate.
Estimate the mass of the crate.

Answer 1

Answer:

Explanation:

Given that the weight of the crate is

M=48kg

Then, weight of boy is

W=mg=48×9.8=470.4N

ΣF = ma ,Along y axis

Since the body is not moving in y direction then, a=0 along y axis

N-W=0

N=W

The normal equals weight of boy

N=470.4N

Then, applying frictional force opposing the boy motion.

Fr=μN

Fr=0.5×470.4

Fr=235.2N

Then, this is the forward force that the boy try using to pull the crate.

Since he can barely move his foot he has not yet overcome the coefficient of static friction.

ΣF = ma. , along x-axis

a along x-axis is 0 since he cannot move his foot, I.e he was not moving

F-Fr= 0

F=Fr=235.2N

The forward force the boy apply is 235.2N on the crate

Also, analysing the crate.

The forward force is now F=235.2N

Then the frictional force =Fr

Then,

ΣF = ma. , along x-axis

a along x-axis is 0 since the crate did not move,

F-Fr=0

Fr=F=235.2N

This is the frictional force on the crate.

Then using frictional law

Fr=μN

N=Fr/μ

N=235.2/0.9

N=261.33N

Now this normal is equal to the weight of the crate

ΣF = ma ,Along y axis

Since the body is not moving in y direction then, a=0 along y axis

N-W=0

W=N=261.33N

Then, since weight is given as

W=mg

m=W/g

m=261.33/9.81

m=26.64kg

The mass of the crate is 26.64kg

Answer 2

Answer:

Mass of the crate = 216 kg

Explanation:

$f = \mu mg$

Since it is desirable that the boy moves the crate without slipping, the static coefficient of the boy will be used.

Since the goal is to push the crate the kinetic coefficient of the crate will be used.

For the boy:

$\mu_{b} = 0.90\\mass, m_{b} =48 kg\\g = 9.8 m/s^{2}$

For the crate:

$\mu_{c} =0.20\\m_{c} =?$

For the boy to be able to successfully push the crate, the net force of the boy and the crate must be zero

$\mu_{b} m_{b} g + (-\mu_{c} m_{c} g)=0\\\mu_{b} m_{b} g = \mu_{c} m_{c} g\\\mu_{b} m_{b}= \mu_{c} m_{c}$

$m_{c} = \mu_{b} m_{b} / \mu_{c} \\m_{c} = (0.9*48)/0.2$

$m_{c} = 216 kg$