They give us the cable tension t = 1500 newton. We assume that the mass of the cable is negligible compared to that of the tower.
We have a force t of 1500 Newton. This force has a vertical component on the y axis and a horizontal component on the x axis.
Of these two components of force, we are especially interested in calculating is the magnitude of the vertical component.
If the angle that it forms with the ground is 50 °, then the vertical component of the force is:
Fy = 1500sin (50)
Fy = 1149 N. <1200 N
The wire will not be loose
Answer:
Explanation:
Let the bigger crate be in touch with the ground which is friction less. In the first case both m₁ and m₂ will move with common acceleration because m₁ is not sliding over m₂.
1 ) Common acceleration a = force / total mass
= 234 / ( 25 +91 )
= 2.017 m s⁻².
2 ) Force on m₁ accelerating it , which is nothing but friction force on it by m₂
= mass x acceleration
= 25 x 2.017
= 50.425 N
The same force will be applied by m₁ on m₂ as friction force which will act in opposite direction.
3 ) Maximum friction force that is possible between m₁ and m₂
= μ_s m₁g
= .79 x 25 x 9.8
= 193.55 N
Acceleration of m₁
= 193 .55 / 25
= 7.742 m s⁻²
This is the common acceleration in case of maximum tension required
So tension in rope
= ( 25 +91 ) x 7.742
= 898 N
4 ) In case of upper crate sliding on m₂ , maximum friction force on m₁
= μ_k m₁g
= .62 x 25 x 9.8
= 151.9 N
Acceleration of m₁
= 151.9 / 25
= 6.076 m s⁻².
The directions arrow<span> is </span>always<span> going the wrong </span>way<span>.</span>
