In earlier days, horses pulled barges down canals in the mannershown in Fig. 5-4
ID: 1669994 • Letter: I
Question
In earlier days, horses pulled barges down canals in the mannershown in Fig. 5-47. Suppose the horse pulls on the rope with aforce of 7800 N at an angle of = 24° to the direction ofmotion of the barge, which is headed straight along the positivedirection of an x axis. The mass of the barge is 9500 kg,and the magnitude of its acceleration is 0.15 m/s2. Whatare (a) the magnitude and (b) the direction of the force on thebarge from the water? (Measure the direction clockwisefrom the direction of motion, and give your angle to the nearestdegree.)
Fig. 5-47
Explanation / Answer
Since this is an equilibrium problem we must make the forces equal in both the horizontal and vertical directions, so in the vertical we have 7800Sin24-FSin=0 and in the horizontal we have 7800Cos24-FCos=(9500)(.15). Now we solve for F in one of the equations to get F=7800Sin24/Sin and then 7800Cos24-(7800Sin24/Sin)Cos=1425 then we have 7800Cos24-7800Sin24/Tan=1425 and we solve to get Tan= .5565 or =29.1 degrees+180 degrees=209.1 deg because the angle should lie in the 3rd quadrant, but this is counterclockwise from the direction of motion, so clockwise would just be 360-209.1 deg=150.9 degrees clockwise. The forces magnitude is then F=7800Sin24/Sin29.1=6523.4 N. (we use 29.1 degrees because this gives us the correct vertical component for second force)