Pulleys and weights - finding acceleration and tension

AI Thread Summary
To find the acceleration of the 24 kg block and the tension in the string, one must apply Newton's second law (F=ma) and consider the relationship between the masses and their movements. The 6 kg block's descent affects the 24 kg block's horizontal movement, with the horizontal distance being half of the vertical distance due to the pulley system. The force exerted by the 6 kg block can be calculated using its weight, which is influenced by gravity, and this force will help determine the acceleration of the larger block. Tension in the string can be found by analyzing the forces acting on both blocks, considering the acceleration and the gravitational force on the 6 kg block. Understanding these relationships is crucial for solving the problem effectively.
iamjohnny56
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Homework Statement


A 24 kg block with a pulley attached slides along a frictionless surface. It is connected by a massless string to a 6 kg block via the arrangement shown.
The acceleration of gravity is 9.81 m/s2 .

The first part wanted you to find the horizontal distance traveled by the 24 kg block if the 6 kg blocked descended 10.1 cm. I found the answer, which is just half of that distance, 5.05 cm.

However, I don't know how to find the other two questions:

1. Find the acceleration of the 24 kg block. Answer in units of m/s2

2. Find the tension in the connecting string. Answer in units of N


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Homework Equations



F=ma

The Attempt at a Solution



I'm not asking for answers - but could someone point me in the direction I should take? I'm not sure about how to start this...
 
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If the 6 kg mass moves twice as far as the 24 kg mass, you get a multiplier of 2 in the force on the larger mass due to the weight of the smaller mass. Remember force times distance equals force times distance.

Realizing that, you can determine the force sliding the 24 kg mass on the table.
 
So if

(Smaller) = (Larger)
Fd = Fd
F(2x) = F(x)

Wouldn't the F cancel out? I'm a bit confused about what you said - could you perhaps elaborate a little bit more?
 
The F's are not the same. One F is twice the other...

This is how block and tackle works - I'm not talking football here.

It's F1 * d1 = F2 * d2. If you move twice as far, you double the force. However, the work done is the same.
 
Think of a little gear turning a larger gear. Torque gets multiplied by the ratio.

If you use a longer pry bar, you can lift more. It's force(1) times distance(1) equals force(2) times distance(2).
 
Must sign off now.
 
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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