Answer:
H = 0.673
Explanation:
given,
side of cubical crate = 0.74
weight of the crate = 600 N
magnitude of force = 330 N
the Horizontal distance of its Center of mass
= 0.74/2
= 0.37
Let the required Height be H
By Balancing the Torques, we get
H x 330 N = 0.37 x 600
330 H = 222
H = 0.673
hence, the height above the floor where force is acting is equal to 0.673 m
Answer:
Hi where is question. Hope you understand me
Answer:
hello your question lacks some data and required diagram
G = 77 GPa, т all = 80 MPa
answer : required diameter = 252.65 * 10-^3 m
Explanation:
Given data :
force ( P ) = 660 -N force
displacement = 15 mm
G = 77 GPa
т all = 80 MPa
i) Determine the required diameter of shaft BC
considering the vertical displacement ( looking at handle DC from free body diagram )
D' = 0.3 sin∅ , where D = 0.015
hence ∅ = 2.8659°
calculate the torque acting at angle ∅ of CD on the shaft BC
Torque = 660 * 0.3 cos∅
= 660 * 0.3 * cos 2.8659 = 198 * -0.9622 = 190.5156 N
hello attached is the remaining part of the solution
Pretty sure it’s false....................