Question

A) 70 kg man s standing in an elevator. Find the reaction  of the elevator on the man if a) the elevator is accelerating upward at 1.5 m/s^2 b) the elevator is accelerating downward at 1.5m/s^2 c) the elevator is descending at a constant velocity 3m/s. B) If the elevator and its contents has a total mass of 1500kg, and it is sup four cables each having a tension T. ? find the tension T IN EACH OF THE CAB of the three cases in Part A. 2.A 2kg block with an initial velocity of 10m/s slides upward a distance plane. it then returns downward to its starting point of which is now 8m/s. If the plane exerts a constant kinetic frictional block, find the distance s. 3.An object initially moving East at 200m/sec explodes into B).Piece A has a mass of 20kg and final velocity of 400m/s( magnitude and direction)?

Answer 1

$\dpi{100} A)$

The force R on the man will be in upward direction .

$\dpi{100} a)$ When a = 1.5 upwards

$\dpi{100} ma = R- mg$

$\dpi{100} R = m(g+a)= 70 (9.8 + 1.5) =\mathbf{ 791 N}$

$\dpi{100} b)$ When a = 1.5 downwards

$\dpi{100} ma = mg - R$

$\dpi{100} R = m(g-a)= 70 (9.8 - 1.5) =\mathbf{ 581 N}$

$\dpi{100} c)$ When velcoity is constant , a = 0

$\dpi{100} ma = R- mg = 0$

$\dpi{100} R = m(g)= 70 (9.8) =\mathbf{ 686 N}$

$\dpi{100} B)$ Similarly , The cables support the system consisting of man and elevator.

So, R = 4T

and M = 70 +1500 = 1570

$\dpi{100} a)$ When a = 1.5 upwards

$\dpi{100} Ma = 4T- Mg$

$\dpi{100} T =\frac{ M(g+a)}{4}= \frac{1570 (9.8 + 1.5)}{4} =\mathbf{ 4435.25 N}$

$\dpi{100} b)$ When a = 1.5 downwards

$\dpi{100} Ma = Mg - 4T$

$\dpi{100} T=\frac{ M(g-a)}{4}= \frac{1570 (9.8 - 1.5)}{4} =\mathbf{ 3257.75 N}$

$\dpi{100} c)$ When velcoity is constant , a = 0

$\dpi{100} Ma = 4T- Mg = 0$

$\dpi{100} T =\frac{ Mg}{4}= \frac{1570 (9.8)}{4} =\mathbf{ 3846.5 N}$

Answer 2

Hello Friend,

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Answer for first question

A.

a. N-m*g=m*a

and we get N = m (g+a) = 791 N

b. N-m*g =-m*a

N = m(g-a)=581 N

c.as velocity is constant , a= 0

that is N = m*g = 686 N

B.

4T