Q2. True or False? A force is a push or a pull that acts on
2 answers:
You might be interested in
The crate moves at constant velocity, this means that its acceleration is zero, so the net force acting on the crate is zero (Newton's second law).
There are only two forces acting on the crate: the force F applied by the worker and the frictional force, acting in the opposite direction:
, where 
is the coefficient of friction and 
is the mass of the crate. Since the net force should be equal to zero, the two forces must have same magnitude, so we have:
And so, this is the force that the worker must apply to the crate.
There are only two forces acting on the crate: the force F applied by the worker and the frictional force, acting in the opposite direction:
And so, this is the force that the worker must apply to the crate.
Answer:
h = 80m L = 60m
Explanation:
Độ cao Ng ta ném hòn bi là
t= => 4^2 =
h= 80m
Tầm bay xa:
L= Vo . t = 15.4 = 60m
Chúc học tốt nghen
Answer:
τ (bc. max) =25.37 MPa
Explanation:
From the question, T = 1.3Kn.m;
(G = 77.2 GPa) and from the image of this solid shaft system i attached;
d(ab) = 50mm; d(bc) = 38mm; L(ab) =0.2m and L(bc) = 0.25m
So ΣT = 0 → Ta + Tc = 1.3Kn.m
So the system is statically indeterminate.
Let's check at the equation that makes it compatible ;
ψ = 0 and ψ(c/b) + ψ(b/a) + ψ(a) = 0
[{T(bc)} / {J(bc)G}] + [{T(ab)} / {J(ab)G}] + 0 =
ΣT = 0 and T(bc) = T(c)
ΣT = 0 and T(ab) = T(c) - 1.3 Kn.m
Now,
[(T(c) x 250mm)/{(π/2)(19^4)}] + [{((T(c) - 1300000 Nmm) x 200mm}/{(π/2)(25^4)}] = 0
So, T(c) = 273374 Nmm = 273.374Nm
T(a) = 1300Nm - 273.374Nm = 1026. 63Nm
From the beginning we saw that;
T(bc) = T(c) = 273.374Nm
Now let's find the maximum shear stress in shaft BC;
τ (bc. max) = {τ (bc) x r(bc)} / J(bc)
τ (bc. max) = (273374Nmm x 19mm)/ {(π/2)(19^4)} = 5194106 / 204707
= 25.37 MPa
