Answer:
<em>The force applied to the car was 30,600 N</em>
Explanation:
According to the second Newton's law, the net force applied to an object of mass m is:
![F_n=m.a\qquad\qquad [1]](https://tex.z-dn.net/?f=F_n%3Dm.a%5Cqquad%5Cqquad%20%5B1%5D)
Where a is the acceleration at which the object moves. The net force can be also calculated as the sum of all forces acting on the body.
We have a car of m=2000 Kg, being accelerated at 5.5 m/s^2 by a force F (unknown) directed upwards.
Considering the force is upwards and the weight of the car (W) is directed downwards, the net force is:
![F_n=F-W\qquad\qquad [2]](https://tex.z-dn.net/?f=F_n%3DF-W%5Cqquad%5Cqquad%20%5B2%5D)
Being W=m.g
Equating [1] and [2]:
Adding W:
Substituting:
F=30,600 N
The force applied to the car was 30,600 N
A force may cause an object to turn about a pivot. The turning effect of a force is called the moment of the force. Moments act about a pivot in a clockwise or anticlockwise direction.
Answer:
The location of the shear center o is 0.033 or 33 m
Explanation:
Solution
Recall that,
The moment of inertia of the section is = I = 0.05 * 0.4 ^3 /12 + 0.005 * 0.2 ^3/12
= 30 * 10 ^ ⁻⁶ m⁴
Now,
The first moment of inertia is
Q =ῩA = [ (0.1 -x) + x/2] (0.005 * x)
= 0.5x * 10 ^⁻³ - 2.5 x * 10⁻³ x²
Thus,
The shear flow is,
q = VQ/I
so,
P = (0.5x * 10 ^⁻³ - 2.5 x * 10⁻³ x²)/ 30 * 10 ^⁻⁶
P = (16.67 x - 83. 33 x²)
The shear force resisted by the shorter web becomes
Vw,₂ = 2∫ = ₀.₁ and ₀ = P (16.67 x - 83. 33 x²) dx = 0.11x
Then,
We take the moment at a point A
∑Mₐ = 0
- ( p * e)- (Vw₂ * 0.3 ) = 0
e = 0.11 p * 0.3/p
which gives us 0.033 m
= 33 m
Therefore the location of the shear center o is 0.033 or 33 m
Note: Kindly find an attached diagram to the question given above as part of the explanation solved with it.
