**preview**has

**blurred**sections. Sign up to view the full version! View Full Document

**Unformatted text preview: **Chapter 5 Homework 4. An adventurous archaeologist crosses between two rock cliffs by slowly going hand over hand along a rope stretched between the cliffs. He stops to rest at the middle of the rope. The rope will break if the tension in it exceeds 2.50 x 10 4 N, and our hero's mass is 90.0 kg. (a) If the angle is 10.0°, find the tension in the rope. θ = Fx 0 = 2Tsinθ mg- = T2cosθ T1cosθ 0 = T mg2sinθ = T2cosθ T1cosθ = . . ° 90 0 kg9 8 ms22sin10 = T2 T1 = 2540 nt (b) What is the smallest value the angle can have if the rope is not to break? θ = 2Tsinθ mg . × = . . 22 50 104 Nsinθ 90 0 kg9 8 ms2 = . ° θ 1 01 5. A picture frame hung against a wall is suspended by two wires attached to its upper corners. If the two wires make the same angle with the vertical, what must this angle be if the tension in each wire is equal to 0.75 of the weight of the frame? (Ignore any friction between the wall and the picture frame.) = . T 0 75W = 2Tcosθ W ( . ) = 2 0 75 W cosθ W = . ° θ 48 2 7. Certain st reets in San Francisco make an angle of 17.5° with the horizontal. What force parallel to the st reet surface is required to keep a loaded 1967 Corvette of mass 1390 kg from rolling down such a st reet? = T Wsinθ = . . ° 1390 kg9 8 ms2sin17 5 = 4096 nt 8. A large wrecking ball is held in place by two light steel cables. If the mass m of the wrecking ball is 4090 kg, what are (a) the tension T B in the cable that makes an angle of 40° with the vertical °= TBsin50 W = . ° TB 4090 kg9 8 ms2sin50 = 52323 nt (b) the tension T A in the horizontal cable? = ° TA TBcos50 = ° 52323 ntcos50 = 33633 nt Page | 2 9. Find the tension in each cord in the figure if the weight of the suspended object is w. (a) = °+ ° TC TAsin30 TBsin45 °= ° TAcos30 TBcos45 Since °= ° sin45 cos45 : = °+ ° W TAsin30 TAcos30 = ( °+ °) TA sin30 cos30 = . TA 0 732W °= ° TAcos30 TBcos45 ( . ) °= ° 0 732W cos30 TBcos45 = . TB 0 897W (b) °= °+ TBcos45 TAsin30 W °= ° TAcos30 TBsin45 Page | 3 °= ° Since sin45 cos45 : = °- ° W TBcos45 TAsin30 = °- ° TAcos30 TAsin30 = °- ° TAcos30 sin30 = . TA 2 73W °= ° TAcos30 TBsin45 ( . ) °= ° 2 73W cos30 TBsin45 = . TB 3 35W 10. A 1130-kg car is held in place by a light cable on a very smooth (frictionless) ramp. The cable makes an angle of31.0° above the surface of the ramp, and the ramp itself rises at 25.0° above the horizontal. (a) Draw a free-body diagram for the car. (b) Find the tension in the cable. = Fx max = °- . ° 0 Tcos31 1130 kg9 8 ms2sin25 = T 5460 N (c) How hard does the surface of the ramp push on the car? Page | 4 = Fy may = + °- . ° n 5460 Nsin31 1130 kg9 8 ms2cos25 = n 7224 N 11. A man pushes on a piano with mass 180 kg so that it slides at constant velocity down a ramp that is inclined at 11.0° above the horizontal floor. Neglect any friction acting on the piano. Calculate the magnitude of the force applied by the man if he pushes (a) parallel to the incline = Fx max = - 0 F mgsinθ = . . °....

View Full Document