A mass of 8 kg is resting on a horizontal, frictionless surface. Force 1 is appl

Question

A mass of 8 kg is resting on a horizontal, frictionless surface. Force 1 is applied to it at 28 degrees below the horizontal, force 2 has a magnitude of 17 N and is applied vertically downward, force 3 has a magnitude of 4 N and is applied vertically upwards, and force 4 has a magnitude of 18 N and is applied in the -x direction to the object. When these forces are applied to the object, the object moves 9 meters in the +x direction and the velocity of the object is 7 m/s in the +x direction. What is the normal force acting on the mass in Newtons? hw4 2.JPG

Explanation / Answer

Using 3rd equation of motio

a= v^2/(2d)

a= 7^2/(2*9)= 2.722 m/s^2

Considering equilibrium of forces along x axis

ma = F1 cos x - F4

8*2.722= F1 cos 28 - 18

F1= 45.05 N

Normal force given by

N= mg+ F2+ F1 sin x -F3

= 8*9.8+ 45.05 sin 28+ 17 -4

= 112.55 N

=======

Commemt in case any doubt.. good luck

Related Questions

Navigate

• Browse (All)
• Browse A
• Subjects

---

---

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.