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Answer the marked (*) questions using complete solutions. 3. A ladder of 5 m long with weight equal to 180 N leans on a frictionless wall. The bottom of the ladder sits on a rough surface, making an angle of 53.1°. What are the normal and friction forces on the base of the ladder? Solution: Refer to the diagram below. We then use the conditions for equilibrium. EF, = F, +(-n2) = 0 and 2F, = n1 + (-180 N) = 0 We get n, = 180 N. We have two equations but three unknowns. To get three equations, we use the equi- librium conditon for the torque. Let's set the reference point at the bottom of the ladder. na 5 m 4 m 180 N n. 53.1° F, 3 m 1 = n;(0) sin(90° – 53.1º) + F,(0) sin(53.1º) + –180N(2.5m) sin(90° – 53.1º) + n¿(5m) sin(53.1°). We get n2 = 67.57 N. Therefore, F, = 67.57 N from the first condition. 4. (Continuation) What is the minimum value of static friction coefficient for the ladder to not slip on the floor? Solution: The static friction F, should not exceed u,nj. Therefore, the minimum static friction coeffi- cient should be F, (H)min = This gives (µ,)min = 0.375. Question 5.* Do items 3 and 4 but this time, a man of weight equal to 700 N stands on top of the ladder. The ladder still does not slip.

Answer the marked (*) questions using complete solutions. 3. A ladder of 5 m long with weight equal to 180 N leans on a frictionless wall. The bottom of the ladder sits on a rough surface, making an angle of 53.1°. What are the normal and friction forces on the base of the ladder? Solution: Refer to the diagram below. We then use the conditions for equilibrium. EF, = F, +(-n2) = 0 and 2F, = n1 + (-180 N) = 0 We get n, = 180 N. We have two equations but three unknowns. To get three equations, we use the equi- librium conditon for the torque. Let's set the reference point at the bottom of the ladder. na 5 m 4 m 180 N n. 53.1° F, 3 m 1 = n;(0) sin(90° – 53.1º) + F,(0) sin(53.1º) + –180N(2.5m) sin(90° – 53.1º) + n¿(5m) sin(53.1°). We get n2 = 67.57 N. Therefore, F, = 67.57 N from the first condition. 4. (Continuation) What is the minimum value of static friction coefficient for the ladder to not slip on the floor? Solution: The static friction F, should not exceed u,nj. Therefore, the minimum static friction coeffi- cient should be F, (H)min = This gives (µ,)min = 0.375. Question 5.* Do items 3 and 4 but this time, a man of weight equal to 700 N stands on top of the ladder. The ladder still does not slip.

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Answer the marked (*) questions using complete solutions. 3. A ladder of 5 m long with weight equal to 180 N leans on a frictionless wall. The bottom of the ladder sits on a rough surface, making an angle of 53.1°. What are the normal and friction forces on the base of the ladder? Solution: Refer to the diagram below. We then use the conditions for equilibrium. EFx = F, + (-n2) = 0 and EF, = n, +(-180 N) = 0 We get n, = 180 N. We have two equations but three unknowns. To get three equations, we use the equi- librium conditon for the torque. Let's set the reference point at the bottom of the ladder. n2 5m 4 m 180 N n. 3 m T = n; (0) sin(90° – 53.1°) + F,(0) sin(53.1°) + –180N(2.5m) sin(90° – 53.1°) + n,(5m) sin(53.1°). We get n2 = 67.57 N. Therefore, F, = 67.57 N from the first condition. 4. (Continuation) What is the minimum value of static friction coefficient for the ladder to not slip on the floor? Solution: The static friction F, should not exceed u,nj. Therefore, the minimum static friction coeffi- cient should be F (H)min = n1 This gives (µ,)min = 0.375. Question 5.* Do items 3 and 4 but this time, a man of weight equal to 700 N stands on top of the ladder. The ladder still does not slip.
Transcribed Image Text:Answer the marked (*) questions using complete solutions. 3. A ladder of 5 m long with weight equal to 180 N leans on a frictionless wall. The bottom of the ladder sits on a rough surface, making an angle of 53.1°. What are the normal and friction forces on the base of the ladder? Solution: Refer to the diagram below. We then use the conditions for equilibrium. EFx = F, + (-n2) = 0 and EF, = n, +(-180 N) = 0 We get n, = 180 N. We have two equations but three unknowns. To get three equations, we use the equi- librium conditon for the torque. Let's set the reference point at the bottom of the ladder. n2 5m 4 m 180 N n. 3 m T = n; (0) sin(90° – 53.1°) + F,(0) sin(53.1°) + –180N(2.5m) sin(90° – 53.1°) + n,(5m) sin(53.1°). We get n2 = 67.57 N. Therefore, F, = 67.57 N from the first condition. 4. (Continuation) What is the minimum value of static friction coefficient for the ladder to not slip on the floor? Solution: The static friction F, should not exceed u,nj. Therefore, the minimum static friction coeffi- cient should be F (H)min = n1 This gives (µ,)min = 0.375. Question 5.* Do items 3 and 4 but this time, a man of weight equal to 700 N stands on top of the ladder. The ladder still does not slip.
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