We don't know Carter, and we don't know where he is or what
he's doing, so I'm taking a big chance speculating on an answer.
I'm going to say that if Carter is pretty much just standing there,
or, let's say, lying on the ground taking a nap, then the force of
the ground acting on him is precisely exactly equal to his weight.
The force tending to lift the load (vertical force) is equal to <u>22.5N.</u>
Why?
Since the boy is pulling a load (150N) with a string inclined at an angle of 30° to the horizontal, the total force will have two components (horizontal and vertical component), but we need to consider the given information about the tension of the string which is equal to 105N.
We can calculate the vertical force using the following formula:
Hence, we can see that <u>the force tending to lift the load</u> off the ground (vertical force) is equal to <u>22.5N.</u>
Have a nice day!
The first thing you should do for this case is to find the horizontal and vertical components of the forces acting on the body.
We have then:
Horizontal = 9-9.2cos (58) = 4.124742769 N.
Vertical = 9.2sin (58) = 7.802042485 N
Then, the resulting net force is:
F = √ ((4.124742769) ^ 2 + (7.802042485) ^ 2) = 8.825268826 N
Then by definition:
F = m * a
Clearing the acceleration:
a = F / m
a = (8.825268826) / (3.0) = 2.941756275 m / s ^ 2
answer:
The magnitude of the body's acceleration is
2.941756275 m / s ^ 2
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