# InClined Plane

a 1kg box on a 30degree frictionless incline is connected to a 3kg box on a horizontal frictionless surface. The pulley is frictionless and massless.
A)if the magnitude of F is 2.3N what is the tension in the connecting cord?
I found all of my formulas
M1 "X" direction-----T-M1g = M1a EQ 1
"Y" direction-----N-M1g = 0 EQ 2
M2 "X" direction-----(T-M2cos30) = M2a EQ 3
"Y"--------------(N-M2gcos60) = 0 EQ 4
im kinda lost..Anyone help?

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i think ive made errors in my equations

someone told me this:
Well, the tension in the connecting cord,
if there is NO other (external) Force, is
(4.9 N)(3kg/4kg) , which is NOT 2.3 N .

So either this set-up is not on Earth's surface
or there's an additional Force (horizontal?)
applied to the (1kg) mass.

If there's an extra Force ("by hand") here,
just add it into the sum of forces!

Could anyone explain where he got the numbers?

In a cord the force is transmited independently of the change of direction. Start identifying the forces acting on this system. The force due to gravity acting on m2 is also accelerates m1. What is the acceleration of m1 and m2? What is the force that is applied on m2? If the net force is 2.3, what is the magnitude of the force that is acting on the system? In fact you have to consider the two mass being only one m1 + m2, and then analyze the force acting on this mass. You know that the net force is 2.3 N (that is the force acting on the system, not m2 alone).

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the net force acting on m2 is 2.3N
It doesnt say if there is any acceleration

By acting on m2 it is acting on the whole system.