Block A in the figure (Figure 1) weighs 62.6 N . The coefficient of static frict

Question

Block A in the figure (Figure 1) weighs 62.6 N . The coefficient of static friction between the block and the surface on which it rests is 0.29. The weight w is 10.5 Nand the system is in equilibrium.

Part A

Find the friction force exerted on block A.

SubmitMy AnswersGive Up

Part B

Find the maximum weight for which the system will remain in equilibrium.

SubmitMy AnswersGive Up

Provide FeedbackContinue

Figure 1 of 1

Block A in the figure (Figure 1) weighs 62.6 N . The coefficient of static friction between the block and the surface on which it rests is 0.29. The weight w is 10.5 Nand the system is in equilibrium.

Part A

Find the friction force exerted on block A.

f =   N  

SubmitMy AnswersGive Up

Part B

Find the maximum weight for which the system will remain in equilibrium.

wmax =   N  

SubmitMy AnswersGive Up

Provide FeedbackContinue

Figure 1 of 1

45.0

Explanation / Answer

given

WA = 62.6 N

W = 10.5 N

theta = 45 degrees.

let T is the tension in the inclined string.

As the system is in equilibrium, net force acting on a particle at cross section must be zero.

use, Fnety = 0

T*sin(45) - W = 0

T = W/sin(45)

= 10.5/sin(45)

= 14.8 N

now use, Fnetx = 0

T*cos(45) - fs = 0

fs = T*cos(45)

= 14.8*cos(45)

= 10.5 N <<<<<<-------------Answer

B) maximum static friction on the block A,

fs_max = mue_s*WA

= 0.29*62.6

= 18.2 N

T_max*cos(45) - fs_max = 0

Tmax = fs_max/cos(45)

= 18.15/cos(45)

= 25.7 N

Wmax = Tmax*sin(45)

= 25.7*sin(45)

= 18.2 N <<<<<<-------------Answer

Related Questions

Navigate

• Browse (All)
• Browse B
• Subjects

---