A horizontal 52-n force is needed to slide a 50-kg box across a flat surface at a constant velocity of 3.5 m/s. What is the coef
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Answer:
The mass of the heaviest box you will be able to move with this applied force = 61.4 kg
Explanation:
From the diagram attached, the forces acting on the box include the weight of the box, applied force on the box, normal reaction of the surface on the box and the Frictional force in the opposite direction to the applied force.
For the box to be able to move, the applied force must have a horizontal component that at least matches the Frictional force between the box and the surface. This is the force balance in the horizontal direction.
Resolving the applied force into horizontal and vertical components,
Fₓ = 750 cos 25° = 679.73 N
Fᵧ = 750 sin 25° = 316.96 N
Doing a force balance in the vertical axis,
N = (mg + 316.96)
Frictional force = μN = μ (mg + 316.96)
μ = 0.74, g = 9.8 m/s²
Frictional force = Fᵧ
μ (mg + 316.96) = 679.73
0.74(9.8m + 316.96) = 679.73
7.252m + 234.5504 = 679.73
7.252m = 679.73 - 234.5504 = 445.1796
m = (445.1796/7.252)
m = 61.4 kg
Hope this Helps!!!
Answer:
Explanation:
L = Initial length of segment = 79 m
T = Normal temperature =
l = Width to be left for expansion
= Coefficient of linear expansion of the material =
= Maximum temperature =
= Change in temperature =
The expression of linear expansion is given by
The expression for the minimum gap distance l the engineers must leave for a track rated at temperature is