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A $ 2.0 kg $ box of cucumber extract is being pulled across a frictionless table by a rope at an angle $ \theta=60° $ (from positive direction of $ x $ axis, we have taken horizontal surface of table as $ x $ axis) The tension in the rope is $ 12N $ and causes the box to slide across the table to the right with an acceleration of $ 3.0 m/s² $ But the direction of net force is along the rope. The direction of acceleration isn't. Why is that?

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    $\begingroup$ I've added the homework-and-exercises tag. In the future, please use this tag on this type of question. $\endgroup$ – user4552 Jan 20 '19 at 15:28
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Let the unit vector in the upward direction be $\hat u$.
If the box is the system with no rope pulling on it then there are two external forces acting on the box:

  • downward gravitational attraction due to the Earth $\vec W = -W \hat u$
  • upward force due to the ground (normal reaction) $\vec N = N \hat u$

Since the box is in static equilibrium

$\vec W + \vec N = \vec 0 \Rightarrow -W + N = 0 \Rightarrow N=W=2g$

Now allow the rope to apply an extra external force on the box $\vec F$.

A common misconception is that the normal reaction force stays the same.
Just to show you that there are times when this is not so assume that the force due to the rope acts vertically upwards thus $\vec F = F \hat u$

If $F<W$ it will still be a static equilibrium situation

$\vec W + \vec N' + \vec F = \vec 0 \Rightarrow -W + N' +F = 0 \Rightarrow N'=W-F=2g-F$

So the magnitude of the new normal reaction force $N'$ has been reduced because of the introduction of a third external force acting on the box.

Not realising this is where, I think, the following misconception has arisen.

But the direction of net force is along the rope.

In this case you cannot just take the static equilibrium situation without the rope and then add another force and assume that the two forces before are still of the same magnitude.
Certainly $W$ stays the same but if the force due to rope has an upward component (ie will be trying to lift the box up) the magnitude of the new normal reaction force $N''$ will be less than $N$.

The free body diagram is on the left and the vector addition of the three external forces (roughly to scale) to give the net force $R$ is shown on the right.

enter image description here

Resolving the applied force $\vec F$ into a component in the up direction $F_{\rm u}$ and a component in the right direction $F_{\rm r}$ and applying Newton's third law

$\vec W + \vec N'' + \vec F =m \vec a \Rightarrow -W \hat u + N'' \hat u + F_{\rm u} \hat u +F_{\rm r} \hat r = m a_{\rm u} \hat u + m a_{\rm r} \hat r$

Which gives two equations

$W = F_{\rm u} +N''$ as the upward acceleraction $a_{\rm u}$ is zero and $F_{\rm r} = m a_{\rm r}$ for the horizontal motion.

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You have to look at all the forces acting on the box, the downward force of gravity (mg), the reaction force of the table, and the vertical and horizontal components of the force exerted by the rope. See the free body diagram of the box below.

The force exerted by the rope has both an upward component ($12 \sin 60=10.4 N$) and a horizontal component ($12 \cos 60 = 6 N$). You will see that the upward component of the force of the rope (10. 4 N) plus the upward reaction force of the table (9.2 N) equals the downward force of gravity (19.6 N), so there is no net vertical component of force and therefore no vertical acceleration.

Since there is no friction opposing the horizontal component of the rope force, the box will accelerate horizontally according to Newton’s second law:

$$a_{x}=\frac{F_{x}}{m}= \frac{6}{2}= 3\frac {m}{s^2}$$

Hope this helps. enter image description here

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  • $\begingroup$ There are three forces in vertical direction. Force of gravitation, normal force and vertical component of tension force. Normal force and gravitational force would balance each other. So there is a net force in vertical direction also (vertical component of tension). The object should accelerate along the direction of tension. But it is accelerating in horizontal direction only. Why is that? $\endgroup$ – Mark fuxerbergstein6 Jan 20 '19 at 8:15
  • $\begingroup$ l didn't get it. Please help $\endgroup$ – Mark fuxerbergstein6 Jan 20 '19 at 8:16
  • $\begingroup$ @Markfuxerbergstein6 See revised answer with diagram. $\endgroup$ – Bob D Jan 20 '19 at 16:52

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