Understanding Interaction Forces: Getting Help with a Frustrating Topic

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Interaction forces can be complex, particularly in scenarios involving static equilibrium, such as a frictionless pulley and ramp system. The key concept is that in static equilibrium, the net force acting on the system is zero, meaning all forces are balanced. A specific example involves calculating the mass of a block on a ramp inclined at 35 degrees, using trigonometry to analyze the forces. Drawing a free body diagram can help visualize these forces and their interactions. Understanding these principles is essential for solving related physics problems effectively.
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Can anyone help me with...

Can someone please help me with interaction forces. I am completely stumped about this its making :mad: me really mad!:smile:



Thanks
 
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bubbleeyes said:
Can someone please help me with interaction forces. I am completely stumped about this its making :mad: me really mad!:smile:



Thanks

You cannot expect to receive any kind of meaningful help when you are THIS vague. Be specific. If this is a question regarding school work, please use the Homework, Coursework, & Textbook Questions forum. Take note of the rules for that forum before posting.

Zz.
 
I just don't get the concept of the whole thing.
Question:
A pulley and ramp are frictionless and the black is in static equilibrium. What is the mass of the block? If the degree is 35 and has 20 kg
 
Draw a free body diagram (similar to the triangle of the ramp...) Use trigonometry to find the forces (you have one leg of the triangle already, and all the angles...).
 
bubbleeyes said:
A pulley and ramp are frictionless and the black is in static equilibrium.

When they say the system is in equilibrium, they're saying that the net force is zero: that is, all the forces cancel each other.

Since it's a static equilibrium, the block and anything else being affected by forces in the system is stationary.
 
Thread 'Variable mass system : water sprayed into a moving container'

Thread 'Variable mass system : water sprayed into a moving container'

Starting with the mass considerations #m(t)# is mass of water #M_{c}# mass of container and #M(t)# mass of total system $$M(t) = M_{C} + m(t)$$ $$\Rightarrow \frac{dM(t)}{dt} = \frac{dm(t)}{dt}$$ $$P_i = Mv + u \, dm$$ $$P_f = (M + dm)(v + dv)$$ $$\Delta P = M \, dv + (v - u) \, dm$$ $$F = \frac{dP}{dt} = M \frac{dv}{dt} + (v - u) \frac{dm}{dt}$$ $$F = u \frac{dm}{dt} = \rho A u^2$$ from conservation of momentum , the cannon recoils with the same force which it applies. $$\quad \frac{dm}{dt}...
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