Advertisements
Advertisements
Question
A steel cable with a radius of 1.5 cm supports a chairlift at a ski area. If the maximum stress is not to exceed 108 N m–2, what is the maximum load the cable can support?
Advertisements
Solution 1
Radius of the steel cable, r = 1.5 cm = 0.015 m
Maximum allowable stress = 108 N m–2
Maximum stress = `"Maximum Force"/"Area of cross section"`
∴Maximum force = Maximum stress × Area of cross-section
= 108 × π (0.015)2
= 7.065 × 104 N
Hence, the cable can support the maximum load of 7.065 × 104 N.
Solution 2
Maximum load = Maximum stress x Cross-sectional area
`= 10^8 Nm^(-2) xx 22/7 xx (1.5 xx 10^(-2) m)^2`
`= 7.07 xx 10^4 N`
RELATED QUESTIONS
A rod of length 1.05 m having negligible mass is supported at its ends by two wires of steel (wire A) and aluminium (wire B) of equal lengths as shown in Figure. The cross-sectional areas of wires A and B are 1.0 mm2 and 2.0 mm2, respectively. At what point along the rod should a mass m be suspended in order to produce (a) equal stresses and (b) equal strains in both steel and aluminium wires.
A mild steel wire of length 1.0 m and cross-sectional area 0.50 × 10–2 cm2 is stretched, well within its elastic limit, horizontally between two pillars. A mass of 100 g is suspended from the mid-point of the wire. Calculate the depression at the midpoint.
Two strips of metal are riveted together at their ends by four rivets, each of diameter 6.0 mm. What is the maximum tension that can be exerted by the riveted strip if the shearing stress on the rivet is not to exceed 6.9 × 107 Pa? Assume that each rivet is to carry one-quarter of the load.
When the skeleton of an elephant and the skeleton of a mouse are prepared in the same size, the bones of the elephant are shown thicker than those of the mouse. Explain why the bones of an elephant are thicker than proportionate. The bones are expected to withstand the stress due to the weight of the animal.
A steel blade placed gently on the surface of water floats on it. If the same blade is kept well inside the water, it sinks. Explain.
The breaking stress of a wire depends on
Modulus of rigidity of ideal liquids is ______.
Modulus of rigidity of ideal liquids is ______.
A rod of length l and negligible mass is suspended at its two ends by two wires of steel (wire A) and aluminium (wire B) of equal lengths (Figure). The cross-sectional areas of wires A and B are 1.0 mm2 and 2.0 mm2, respectively.
(YAl = 70 × 109 Nm−2 and Ysteel = 200 × 109 Nm–2)
- Mass m should be suspended close to wire A to have equal stresses in both the wires.
- Mass m should be suspended close to B to have equal stresses in both the wires.
- Mass m should be suspended at the middle of the wires to have equal stresses in both the wires.
- Mass m should be suspended close to wire A to have equal strain in both wires.
The value of tension in a long thin metal wire has been changed from T1 to T2. The lengths of the metal wire at two different values of tension T1 and T2 are l1 and l2 respectively. The actual length of the metal wire is ______.
A steel wire having a radius of 2.0 mm, carrying a load of 4 kg, is hanging from a ceiling. Given that g = 3.1πms-2, what will be the tensile stress that would be developed in the wire?
A body of mass m = 10 kg is attached to one end of a wire of length 0.3 m. The maximum angular speed (in rad s-1) with which it can be rotated about its other end in the space station is (Breaking stress of wire = 4.8 × 107 Nm-2 and the area of cross-section of the wire = 10-2 cm2) is ______.
The area of the cross-section of the rope used to lift a load by a crane is 2.5 × 10-4m2. The maximum lifting capacity of the crane is 10 metric tons. To increase the lifting capacity of the crane to 25 metric tons, the required area of cross-section of the rope should be ______.
(take g = 10 ms-2)
The stress-strain graph of a material is shown in the figure. The region in which the material is elastic is ______.
Strain is defined as the ratio of:
When an elastic body is in equilibrium in its altered shape, the magnitude of the external deforming force is:
