The only forces acting on a 2.7 kg mass are
~ F1=F1x ˆi−F1y ^j and F2=F2x ˆi+F2y ˆj, where
F1x = 2 N, F1y = 7 N, F2x = 6 N, F2y = 2 N.
What is the magnitude a of the acceleration
of the particle?
Answer in units of m/s2

Question

The only forces acting on a 2.7 kg mass are
~ F1=F1x ˆi−F1y ^j and F2=F2x ˆi+F2y ˆj, where
F1x = 2 N, F1y = 7 N, F2x = 6 N, F2y = 2 N.
What is the magnitude a of the acceleration
of the particle?
Answer in units of m/s2

ghazi · Answer

can you find the resultant force?

anonymous · Answer

by doing cos and sin for each?

ghazi · Answer

no, by using vectors

anonymous · Answer

Do i solve for the x and y component for each?

ghazi · Answer

$$F _{1}=(2 i-7j ) N$$ and $$F _{2}=(6i+2j) N$$ now resultant vector is $$F=F _{1}+F _{2}=(2i-7j)+(6i+2j)=(8i-5j) N$$ 
and magnitude is of $$F _{R}=\sqrt{8^2+5^2}=\sqrt{64+25}=\sqrt{89}=9.43N$$ therefore acceleration = 9.43/2.7 m/s^2

ghazi · Answer

@ncarrillo52  does that help?

anonymous · Answer

yes thank you!!

ghazi · Answer

:)

anonymous · Answer

Consider the three blocks connected by massless inextensible cords over massless frictionless pulleys as shown in the ﬁgure below. The
coeﬃcient of static friction between the (middle) block and the table top is µs = 0.54 and
the system is in equilibrium.
The acceleration of gravity is 9.8 m/s
|dw:1350338054468:dw|

SOLVE THE MAX WEIGHT M COULD BE.

anonymous · Answer

NVM I figured that one out. i forgot to time friction by normal.

ghazi · Answer

friction force= u*mass*acceleration due to gravity