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Chapter 12: Q57P (page 351)

In Fig 12-66, asphere is supported on a frictionless plane inclined at angle θ=45°from the horizontal. Angle ϕis 25°. Calculate the tension in the cable.

Short Answer

Expert verified

Tension in the cable is76 N.

Step by step solution

01

Listing the given quantities

θ=45°

ϕ=25°

Mass of spherem=10 kg

02

Understanding the concept of force and tension

We draw the free body diagram for the sphere. From this, we take the summation of all forces in the direction parallel to the ramp. After solving the equation, we will get the tension in the cable.

Equation:

∑Fx=0

03

Free Body Diagram

04

Calculation of tension in the cable

∑Falongtheinclinedplane=0(Tcosϕ)−(mgsinθ)=0(Tcosϕ)=(mgsinθ)T=(mgsinθ)cosϕT=10 kg×9.8 m/s2×sin45°cos25°T=76 N

Tension in the cable is76 N

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Most popular questions from this chapter

Question: In Fig. 12-26, a uniform sphere of mass m = 0.85 m and radius r = 4.2 mis held in place by a massless rope attached to a frictionless wall a distance L = 8.0 cm above the center of the sphere. Find (a) the tension in the rope and (b) the force on the sphere from the wall.

In Fig. 12-63, a rectangular slab of slate rests on a bedrock surface inclined at angle θ=26°. The slab has length L=43 m, thickness T=2.5 m, and width,W=12 mand 1.0 cm3of it has a mass of 3.2g. The coefficient of static friction between slab and bedrock is 0.39. (a) Calculate the component of the gravitational force on the slab parallel to the bedrock surface. (b) Calculate the magnitude of the static frictional force on the slab. By comparing (a) and (b), you can see that the slab is in danger of sliding. This is prevented only by chance protrusions of bedrock. (c) To stabilize the slab, bolts are to be driven perpendicular to the bedrock surface (two bolts are shown). If each bolt has a cross-sectional area of 6.4 cm2and will snap under a shearing stress of, 3.6×108N/m2what is the minimum number of bolts needed? Assume that the bolts do not affect the normal force.

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Figure:

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