Can someone help me with this physics problem?

Question

A mail bag with mass 120kg is suspended by a vertical rope 6.0 m long. What horizontal force is necessary to hold the bag in a position displaced sideways 3.0 m from its initial position?

so far, i have come up with the equation of F=mgsin(x) but that is wrong, no angle is in the problem.

thanks for any help!

Answer 1

You do have a geometry in this problem. The geometry is a rope that forms an angle with the vertical. Draw a triangle. The hypoteneuse of the triangle is the length of the rope and is 6.0m. The opposite side from the angle formed by the rope and the vertical is 3.0. From this, you can figure out the angle from vertical to be theta=arcsin(3/6)=0.534 radians or 30 degrees.

Now you balance forces. Force from gravity will act downward at the end of the rope and is equal to F_mailweight=120kg*9.8m/s^2=1176N in magnitude. The only vertical force to balance this is the tension in the rope. So you figure out the tension in the rope T=F_mailweight/cos(theta). You now have tension in the rope. The only thing left to do is to figure out what horizontal force that must be applied to counteract the horizonal component of the tension in the rope.

F_horizontal=T*sin(theta)=F_mailweight*tan(theta)

Now you have the force that must be applied horizontally in order to hold this mail bag out stationary at 3m side ways from a 6m rope. Put numbers into the equations and punch it through the calculator and you'll get the number.

Answer 2

you would have to view this sideways, bag to the right, to get the geometry. Then the displacement 3 m gives a sine of (3 m divided by radius 6 m (rope length)) so the sine is .5

So the force is 120kg *.5 = 60 kg to hold it out 3 M.