Question

The 250-lb block shown in has impending motion up the plane caused
by the horizontal force of 470 lb. Determine the coefficient of static
friction between the contact surfaces.

Answer 1

Answer:

Explanation:

. इनमें प्रथम प्रेरणात्मक क्रिया रूप है-

बी) चडाई

ए) लिखावट

सी) सुनाना

डी) बचना

​

Answer 2

Explanation:

Given The 250-lb block shown  has impending motion up the plane caused by the horizontal force of 470 lb. Determine the coefficient of static friction between the contact surfaces.

• So the block has impending motion up the plane. We need to find the coefficient of static friction between the forces.
• So we have
•      ∑ Fy = 0
• N = 470 sin 30 + 250 cos 30
•   = 235 + 125 x 1.732
•  = 235 + 216.5
• N = 451.5 lb
•   ∑ Fx = 0
• f + 250 sin 30 = 470 cos 30
• f + 125 = 407.02
• f = 407.02 – 125
• f = 282.02 lb
• Now we have
•                                          f = μ N
•                                     282.02 = μ (451.5)
•                                        μ = 282.02 / 451.5
•                                         μ = 0.62

Reference link will be

https://brainly.in/question/12349780