Physics' Quesion

Question

Pulling a Crate A worker drags a crate across a factory floor by pulling on a rope tied to the crate (Fig. 6-60). The worker exerts a force of 450 N on the rope, which is inclined at 38° to the horizontal, and the floor exerts a horizontal force of 125 N that opposes the motion.

Figure 6-60 (image is in this link http://www.webassign.net/hrw/05_44.gif )

(a) Calculate the magnitude of the acceleration of the crate if its mass is 315 kg.
        

(b) Calculate the acceleration if the crate's weight is 315 N.

                

Answer 1

We can start this question by figuring out our horizontal force components because that's where the acceleration is, meaning the crate will be moving in a horizontal only motion. To do this we need to draw a free body diagram for the crate:

http://www.physicsclassroom.com/class/newtlaws/u2l2c6.gif

Since we're only looking for horizontal components we only need to worry about the force of friction and the force being applied in the horizontal direction. The force of gravity will be needed when we calculate the acceleration for the different weights of the crate. 

The first step is to figure out the applied force on the horizontal. We can do this by taking the cos of 38º and multiplying that by 450N. The reason for this is because if we draw a straight line down from where the worker's hand is we get a right triangle and to figure out the horizontal leg we use SohCahToa where 450N is the hypothenuse and our angle is 38. 

450N*cos(38º)=354.6N

So we now have two horizontal forces: a force of 125N going left and 354.6N going right. Let's now set the force of friction as a negative force and the applied force as positive. The difference in these two forces will give us a total applied force of:

354.6N-125N= 229.6N 

Now that we have a force of 229.6N pulling on the crate we can calculate the acceleration. 

(a) Remember that a N is actually kg*m/s². So we need to cancel the kg out. In this case that's easy because the weight of the crate is already in kg and all we need to do is divide the force by the weight:

229.6N/315kg=.729m/s². 

It's always a good idea to write the units down to keep track of whether you have the right figure or not. 

(b) In this case our weight is in N. To get to kg we need to divide out the m/s² part (acceleration). Well the way that's done is by using acceleration due to gravity which is 9.81m/s² (recall this is what we needed from the beginning). So we now have:

315N/9.81m/s²=32.1kg

Now that we have our crate's weight we can simply divide the force applied on it by the weight to get acceleration:

229.6N/32.1kg=7.15m/s²

If you have any more questions feel free to post. 

 

Answer 2

You must draw a force diagram on the box. 

• The weight acts down and is perpendicular to the movement.  It doesn't do any "work"
• The Floor acts horizontally, opposing motion
• Normally, I would consider the friction to be proportional to the Normal force and do some calculations to figure out the normal force, but here they just tell you that there is opposition to motion that acts horizontally, so let's go with that...
• The man acts at an angle, but only the portion of the force he exerts horizontally does any "work" to make the crate move...so we have to figure out the horizontal part of his force.
• A triangle (vector) diagram would be set up so that the man's force is the hypothenuse, and the horizontal component is adjacent to the angle (38 degrees).  The vertical component is the opposite leg.
• If you remember your triangle trigonometry, this means the the magnitude of the horizontal vector is F*(cos θ)

Our balanced force equation is then:

M*a = F*(cos θ) - f,

• where f is the resistance from the floor

(a)         a = {F*(cos θ) - f} / M

              ={450*(0.788)-125}/315

              =0.729 meters/s²

(b)   if the weight is 315 N, then the mass is 315 = M*(9.81)

      solve for M = 32.11kg

and plug this in to the equation

a = {450*(0.788)-125}/32.1 = 7.15 m/s²