Answer:
a) γ =0.055556
b) t = 0.4 MPa
Explanation:
Given:
- The dimensions of rubber block : 18 x 21 x 25
- A load was applied at upper frame P = 420 N
- The rubber deflects dx = 1 mm downwards
Find:
(a) average shear strain in the rubber mounts
(b) average shear stress in the rubber mounts.
Solution:
- For average shear strain we have the definition:
γ = dx / y
Where,
γ: The shear strain.
dx : Deflection along the shear force
y : The length perpendicular to deflection.
- From given data we have dx = 1mm, and the dimension of block perpendicular to deflection is the a dimension. Hence, dx = 0.001 and y =0.018 m:
γ = 0.001 / 0.018 = 0.055556
- The average shear stress along the mating flat surface. We have from definition:
t = F_shear / Area
- Where, F_shear: The shear force on each rubber block is P/2.
Hence,
t = (P/2) / b*c
Plug values in:
t = (420/2) / (0.021*0.025)
t = 0.4 MPa
Answer:
0.011 N-m
Explanation:
Given that
The mass of a solid cylinder, m = 30 kg
The radius of the cylinder, r = 0.18 m
The acceleration of the cylinder,
It rotates about an axis through its center. We need to find the torque acting on the cylinder. The formula for the torque is given by :
Where
I is the moment of inertia of the cylinder,
For cylinder,
So,
So, the required torque on the cylinder is 0.011 N-m.
Answer : The acceleration of the elevator is,
Explanation :
Formula used to calculate the acceleration of the elevator is:
where,
F = force = 844 M
m = mass of man = 70 kg
g = acceleration due to gravity =
a = acceleration of the elevator = ?
Now put all the given values in the above formula, we get:
Thus, the acceleration of the elevator is,
It is TRUE. The pivot point of a lever is called the fulcrum .