Explore BrainMass

Tension - Equilibrium and Elasticity: Five problems

The graphs are attached in word document.

1. The system in the picture is in equilibruim with the string in the center exactly horizontal. Find (a) tension T 1 , (b) tension T 2 (c) tension T 3 (d) angel pheta

2. The system in the picture is in equilibrium. A concrete block of mass 225 kg hanfs from the end of the uniform strut whose mass is 45.0 kg. Find (a) the tension T in the cable and the (b) horizontal and (c) vertical force components on the strut from the hinge.

3. For the step ladder shown in the picture. Sides AC and CE are each 2.44 m long and hinged at C. Bar BD is a tie-rod 0.762 m long , halfway up. A man weighing 854 Newtons climbs 1.80 m along the ladder assuming that the floor is frictionless and neglect ing the mass of the ladder, find (a) the tension in the tie-rod and the magnitudes of the forces on the ladder from the floor at (b) A and (c) E

4. A construction worker attempts to lift a uniform beam off the floor and raise it to a vertical position. The beam is 2.5 m long and weighs 500 N. At a certain instant the worker hold the beam momentarily at rest with one end 1.5m off the floor as shown in the figure below by exerting a force vector P on the beam, perpendicular to the beam. (a) what's the magnitude of the force exerted by the worker ? (b) What is the magnitude of the (net) force of the floor on the beam? (c) What is the minimum value that the coefficient of static friction between the beam and the floor can have in order for the beam not to slip at this instant ?

5. The picture shows a uniform ramp between two buildings that allows for motion between the buildings due to strong winds. At its left end it is hinged to the building wall, at its right end it has a roller that can roll along the building wall. There is no vertical force on the roller from the building, only a horizontal force with magnitude Fh . The horizontal distance between the buildings is D=4.00 m . The rise of the ramp is h= 0.490m. A man walks across the ramp from the left Picture 5-1 gives Fh as a function of horizontal distance x of the man from the building at the left. What are the masses of (a) the ramp and (b) the man ?


Solution Preview

1. There is typo, in two strings you have written T2, hence, I've considered T3 in horizontal string and T2 in the string inclined at Q (read as theta) with respect to vertical.

For whole system:
Equlibrium in vertical direction: sum(Fv) = 0
T1*cos(35) + T2*cos(Q) = 40+50 = 90 ....(1)
(Here Q read as theta)
In horizontal direction:
538.516sin(35)-T2*sin(Q) = 0 ......(2)
At the junction of T1, T3 (horizontal st
ring) and 40 N, by sine law:
T1/sin(90) = T3/sin(180-35) = 40/sin(90+35)
=> T1 = T3*1.74 = 40*1.22 = 48.8
=> T1 = 48.8 N --Answer
T3 = 48.8/1.74 = 28.05 N (horizontal string) --Answer
From equation 2:
T2*sin(Q) = T1*sin(35) = 48.8*0.574 = 28.01 ....(3)
From ...

Solution Summary

The detailed explanations and calculations solve the problems.