Solving Newton's 2nd Law Problems: Blocks & Crate

[Miagi]

Hi,
I'm having trouble with these problems dealing with Newton's 2nd Law

1] Two blocks are in contact on a frictionless table. A horizontal force is applied to the larger block, as shown in Fig. 5-50. (a) If m1 = 2.0 kg, m2 = 1.4 kg, and F = 2.7 N, find the magnitude of the force between the two blocks. (b) Assume instead that a force of the same magnitude F is applied to the smaller block but in the opposite direction, and calculate the magnitude of the force between the blocks. (Why is the value calculated in (b) not the same as that calculated in (a)?)

So I have no idea how to setup the F=m*a for this problem. I think I need to find another force for the smaller block, but I'm not sure how. Can someone help me.


2] A worker drags a crate across a factory floor by pulling on a rope tied to the crate. The worker exerts a force of magnitude F = 440 N on the rope, which is inclined at angle θ = 42° to the horizontal, and the floor exerts a horizontal force of magnitude f = 110 N that opposes the motion. Calculate the magnitude of the acceleration of the crate if (a) its mass is 370 kg and (b) its weight is 370 N.

For this problem. I tried setting up the F=m*a.
440 - 110 - m_1*g*sin(theta). I got m_1*g*sin(theta) because since it is at an angle, the gravitational for is not directly opposite of the Normal force.
Well I tried this setup but doesn't seem to give me the correct answer.
Can someone help?

Thanks

 

[cepheid]

Hi,
I'm having trouble with these problems dealing with Newton's 2nd Law

1] Two blocks are in contact on a frictionless table. A horizontal force is applied to the larger block, as shown in Fig. 5-50. (a) If m1 = 2.0 kg, m2 = 1.4 kg, and F = 2.7 N, find the magnitude of the force between the two blocks. (b) Assume instead that a force of the same magnitude F is applied to the smaller block but in the opposite direction, and calculate the magnitude of the force between the blocks. (Why is the value calculated in (b) not the same as that calculated in (a)?)

So I have no idea how to setup the F=m*a for this problem. I think I need to find another force for the smaller block, but I'm not sure how. Can someone help me.

Please use the TEMPLATE provided for homework problems from now on (the one you deleted when you typed your post). What have you done so far on this problem?

I would recommend starting by drawing a free-body diagram for each block, indicating all of the relevant forces. This will help you figure out what's going on for each individual block.

You know that the acceleration of the blocks has to be the same, and you can figure out what it is by applying Newton's second law to the combined mass of the two blocks.

 

[SammyS]

As cepheid recommended , draw a free-body diagram !

 

[ileacus]

For 2) the (frictional) force exerted by the floor is given independent of the angle of the rope. The force the worker applies on the rope is given at an angle to the floor. Only part of the force the worker is exerting is used to move the box horizontally. What fraction of the worker's total force is applied horizontally?

 

[rwisz]

,


Hi there,

It seems like you are struggling with applying Newton's second law to these problems. Let's break down the steps to solve these problems and see if that helps.

1) For the first problem, you are given the masses of the two blocks and the applied force. To find the force between the blocks, you can use the formula F=ma. However, since the blocks are in contact and there is no friction, the acceleration of the two blocks must be the same. So we can set up two equations, one for each block:

For the larger block: F=m1a
For the smaller block: F=m2a

Since we want to find the force between the blocks, we can set these two equations equal to each other and solve for a:

m1a = m2a
a = F/(m1+m2)

Now that we have the acceleration, we can use the formula F=ma again to find the force between the blocks:

F = (m1+m2)a
F = (2.0+1.4)(2.7)
F = 8.1 N

For part (b) of the problem, we have to consider the fact that the force is now acting in the opposite direction on the smaller block. This means that the acceleration of the smaller block will be in the opposite direction as well. So when we set up the equations, we have:

For the larger block: F=m1a
For the smaller block: -F=m2a

Solving for a, we get:
a = -F/(m1+m2)

Plugging in the given values, we get:
a = -2.7/(2.0+1.4)
a = -0.675 m/s^2

Since the acceleration is negative, it means that the smaller block is moving in the opposite direction as the applied force. And since the force is acting in the opposite direction, the magnitude of the force between the blocks will be different than in part (a).

2) For the second problem, it seems like you are on the right track with setting up the equation F=ma. However, there are a few things to consider. First, the force of 110 N that is opposing the motion is not a force of gravity, it is a force from the floor. So it should not be included in the gravitational force term m1g*sin(theta). Instead,