Question

A 1000-kg barge is being towed by means of two horizontal cables. One cable is pulling with a force of 80.0 N in a direction 30.0° west of north. In what direction should the second cable pull so that the barge will accelerate northward, if the force exerted by the cable is 120 N? Assume that the water exerts no appreciable frictional drag on the barge.

Answer 1

Answer:

$19.5^{\circ}$E of N

Explanation:

Mass of barge=m=1000 kg

$F_1=80 N$

$\theta_1=30^{\circ}$

$F_2=120 N$

We have to find the direction in which the second cable should pull.

In north direction, the force applied=$F_N=F_1cos\theta_1=80cos30^{\circ}=40\sqrt 3 N$

In west direction , the force applied=$F_W=F_1sin\theta_1=80sin30^{\circ}=40 N$

The force applied in east direction is equal and opposite to force applied in west direction.

Therefore, $F_2sin\theta=40$

$120sin\theta=40$

$sin\theta=\frac{40}{120}=\frac{1}{3}=0.333$

$\theta=sin^{-1}(0.333)=19.5^{\circ}$

Hence, the direction in which the second cable should pull=19.5 degree E of N