Solving Pulley Type Problem: Mass, Forces & Momentum

[nysnacc]

Homework Statement

Homework Equations

Sum of Force
g = 32.174 ft/ s^2

The Attempt at a Solution

I consider the right hand side weight,

T = -mg = - 10*32.174 = 321.74 lb
Why is the answer 3.33, which seems to be even smaller than the mass??
Should I consider the momentum, or something with the 5 lb mass??

Thanks.

 

[kuruman]

The tension is equal to the weight only if the velocity of the mass is constant. Here there is acceleration and you need to take that into account.

 

[kuruman]

Also check your units. Is 10 lb x 32.174 ft/s² really equal to 32.174 lb?

 

[nysnacc]

I see,

Also check your units. Is 10 lb x 32.174 ft/s² really equal to 32.174 lb?

I see (not familiar with US Unit) my thought was T = mg , so if it is metric unit T (N) = m (kg) * g (m/s^2)
so in this case maybe T (lb*ft/s)

And how do you find the acceleration? thanks

 

[kuruman]

First the units. When you see a number with lb. next to it, that means weight which is a force. To find the mass (the unit is slugs) you divide the weight in lb. by 32.174 ft/s². The number you get should be used wherever "m" appears. If g appears next to m, as in mg, you just use the number for weight in lb.

Now for the acceleration. Newton's Second law says that the um of all the forces is equal to mass times acceleration.
1. What is the sum of all the forces acting on the mass on the table?
2. What is the sum of all the forces on the hanging mass?

Once you have the answer to these two questions, you can set each equal to the appropriate mass times the acceleration.

 

[jack action]

About the US units:

'lb' is already a unit of force. In the US system, we measure a weight in 'lb' and in the SI system we measure a mass in 'kg'. And - of course - weight = mass X g, in either system. The weight in the SI System has the basic unit of Newton and the basic unit for mass in the US system is the slug.

By definition, 1 lb_f = 1 slug.ft/s² (note subscript 'f' for 'pound-force'). So the correct way to use $W=mg$ with your numbers is to find the mass in slug which is 10 lb_f / (32.2 ft/s²) = 0.31056 slug.

By definition, 1 lb_f = 32.2 lb_m.ft/s² (note subscript 'm' for 'pound-mass'). This means that 1 slug = 32.2 lb_m, such that 0.31056 slug X 32.2 lb_m/slug = 10 lb_m. So you can see that 10 lb_f is equivalent to 10 lb_m under the Earth acceleration. This way, there is no need to do a conversion (at least numerically).

 

[nysnacc]

First the units. When you see a number with lb. next to it, that means weight which is a force. To find the mass (the unit is slugs) you divide the weight in lb. by 32.174 ft/s². The number you get should be used wherever "m" appears. If g appears next to m, as in mg, you just use the number for weight in lb.

Now for the acceleration. Newton's Second law says that the um of all the forces is equal to mass times acceleration.
1. What is the sum of all the forces acting on the mass on the table?
2. What is the sum of all the forces on the hanging mass?

Once you have the answer to these two questions, you can set each equal to the appropriate mass times the acceleration.

1. x direction: Tension y direction: Normal force, and Weight
2. x direction: none y direction: Tension and weight

So I set F_x1 = F_x2?
and same for y dir?

 

[kuruman]

Consider the mass on the table first. There is only one force in the x-direction which means that the net force in the x-direction is just T. So, according to Newton's Second law, in the horizontal direction
1. T = ??
In the vertical direction, the mass just slides on the table without jumping off it or sinking into it. What does that make the vertical acceleration equal to?
2. a_vertical= ??

Now for the hanging mass. Nothing is going on in the horizontal direction. What is the net force in the vertical direction?
3. F_net,y= ??

Finally, set the net force in each direction equal to the (appropriate) mass times acceleration in the same direction. You also need to consider whether the two masses have the same acceleration and why or why not.