(a) You pull a box with a constant force across a frictionless table using an attached rope held horizontally. If you now pull the rope with the same force at an angle to the horizontal (with the box remaining flat on the table), does the acceleration of the box increase, decrease, or remain the same? Explain. (b) What if there is friction?

According to Newton’s second law, the force acting on an object is directly related to the object’s mass and acceleration. For a higher-mass object, a higher value of force is required to accelerate it.

Part (a)

Here, $F$is the force that acts on the box, $N$is the normal force on the box, $m$is the mass of the box, $a$is the acceleration of the box, $\theta$is the angle made by the force with the horizontal, $F\cos \theta$is the horizontal component of the force, and $F\sin \theta$is the vertical component of the force.

The expression for the force that acts on the box when it is pulled parallel to the horizontal can be written as

$F=ma\dots \left(i\right)$

The horizontal force that acts on the box when the force is at inclination is given by

$F\cos \theta$.

The angle $\left(\theta \right)$varies from 0 to 1 for cosine and sine angles. Then, the limits for $\cos \theta$will be

$0<\cos \theta <1$.

Multiplying the above relation with $F$,

$\begin{array}{c}F\left(0\right)<\left(F\right)\cos \theta <F\left(1\right)\\ 0<F\cos \theta <F\end{array}$.

Substituting the value of equation (i) in the above equation,

role="math" localid="1645082333072" $\begin{array}{c}0<F\cos \theta <ma\\ a>\frac{F\cos \theta }{m}\dots \left(ii\right)\end{array}$

Consider that the acceleration of the box is $a\text{'}$when it is pulled at an inclination.

Then, the horizontal force that acts on the force will be

$F=ma\text{'}$.

Substituting this value in equation (ii),

$\begin{array}{c}a>\frac{ma\text{'}\cos \theta }{m}\\ a>a\text{'}\cos \theta \end{array}$.

Thus, from the above relation, it is clear that theacceleration of the box decreases if the box is pulled at an angle.

Part (b)

Here, $F$is the force that acts on the box, $N$is the normal force on the box, $m$is the mass of the box, $a$is the acceleration of the box, $\theta$is the angle made by the tension force with the horizontal, $T\cos \theta$is the horizontal component of the tension force,$T\sin \theta$is the vertical component of the tension force, and $f$is the frictional force.

If there is no frictional force existing between the block and the table, the expression for the net force can be written as

$\begin{array}{c}F=ma\\ T=ma\\ a=\frac{T}{m}\dots \left(iii\right)\end{array}$

Consider that $a\text{'}$is the acceleration of the box when there is a frictional force existing between the block and the table.

If there is a frictional force existing between the block and the table, the expression for the net force can be written as

role="math" localid="1645082487698" $\begin{array}{c}ma\text{'}=T\cos \theta -f\\ a\text{'}=\frac{T\cos \theta -f}{m}\dots \left(iv\right)\end{array}$

The condition for $\cos \theta$is given by

$0<\cos \theta <1.$

The above expression for the tension force can be written as

$0<T\cos \theta <T$… (v)

Thus, from equations (iii), (iv), and (v), it is clear that theacceleration of the box decreases if there is a frictional force existing between the table and the box.