Em um centro de distribuição uma empilhadeira carrega uma caixa de 1000kg até o caminhão responsável pelo transporte desta carga
1 answer:
Answer:
F = 592238.09 N
Explanation:
To find the mean force you first calculate the acceleration by using the following kinematic equation:
you do "a" the subject of the equation and replace the values of the other parameters:
next, the force, by using the second Newton law is:
hence, the force is 592238.09 N
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TRANSLATION:
Para encontrar a força média, primeiro calcule a aceleração usando a seguinte equação cinemática:
você faz "a" o assunto da equação e substitui os valores dos outros parâmetros:
Em seguida, a força, usando a segunda lei de Newton, é:
portanto, a força é 592238.09 N
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Answer:
The question is incomplete, below is the complete question "A particle moves through an xyz coordinate system while a force acts on it. When the particle has the position vector r with arrow = (2.00 m)i hat − (3.00 m)j + (2.00 m)k, the force is F with arrow = Fxi hat + (7.00 N)j − (5.00 N)k and the corresponding torque about the origin is vector tau = (4 N · m)i hat + (10 N · m)j + (11N · m)k.
Determine Fx."
Explanation:
We asked to determine the "x" component of the applied force. To do this, we need to write out the expression for the torque in the in vector representation.
torque=cross product of force and position . mathematically this can be express as
Where
and the position vector
using the determinant method to expand the cross product in order to determine the torque we have
by expanding we arrive at
since we have determine the vector value of the toque, we now compare with the torque value given in the question
if we directly compare the j coordinate we have
0 = u^2 + 2*(-2)*9600
u^2 = 38400
u = 195.96 m/s