Respuesta :
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:
Mass of the crate = 216 kg
Explanation:
[tex]f = \mu mg[/tex]
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:
[tex]\mu_{b} = 0.90\\mass, m_{b} =48 kg\\g = 9.8 m/s^{2}[/tex]
For the crate:
[tex]\mu_{c} =0.20\\m_{c} =?[/tex]
For the boy to be able to successfully push the crate, the net force of the boy and the crate must be zero
[tex]\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} \\[/tex]
[tex]m_{c} = \mu_{b} m_{b} / \mu_{c} \\m_{c} = (0.9*48)/0.2[/tex]
[tex]m_{c} = 216 kg[/tex]
