Rotational force

Hello,

My question is this:
A mail bag with a mass of 105 kg is suspended by a vertical rope of length 7.00 m. What horizontal force is necessary to hold the bag in a position displaced sideways a distance of 1.00 m from its initial position?

I tried to start this problem but I keep getting stuck. I start out figuring that it is an isosceles triganle with the legs being 7m and the base being 1 m. from there i figured that if i resolved the 1 meter into its horizontal and vertical components, I would be able to find the force using Newtons second law using 9.8 meters per second squared as the acceleration and the mass as 105 kg. but i think that using 9.8 meters per second squared is wrong because that is acceleration due to gravity and I only need horizontal acceleration....... I am confusing myself with this problem.......

 

thanks for the welcome!!!

so if i do the diagram, i have two ropes, one hanging vertically (that is slanted) and one pulling horizontally, and I am calculating the tension on the one pulling horizontally? how do i know the angles at which the semi vertical rope is slanted?am i on th right track? Thanks for your reply =)

 

thank you. I have come to the answer of 149 Newtons. The second part of the problem says to calculate the work done by the worker to get to this postion. I have tried work= 149N*1m and that does not work, then w= 149n*1m*cos (82 degrees), and that does not work. is ther something i am missing?