QUESTION 6 One 7.93 kg paint bucket is hanging by a massless cord from another 7

Question

QUESTION 6
One 7.93 kg paint bucket is hanging by a massless cord from another 7.72 kg paint bucket, also hanging by a massless cord, as shown in Fig. 4-44.
If the buckets are pulled up by an acceleration of 1.03 m/s2 by the upper chord, what is the tension in the top cord?
A force of 45.24N is required to start a 7.28kg box moving across a horizontal floor. What is the coefficient of static friction between the box and the floor?
QUESTION 10
A force of 46.61 N is required to start a 5.57 kg box moving across a horizontal floor.
If the 46.61 N force continues, the box accelerates at 0.68 m/s2. What is the coefficient of kinetic friction?
QUESTION 11
A 12.21kg box is released on a 31° incline and accelerates down the incline at 0.35m/s2. Find the friction force impeding it's motion.
QUESTION 12
A 13.27kg box is released on a 33° incline and accelerates down the incline at 0.21m/s2. What is the coefficient of kinetic friction?
QUESTION 14
Suppose two boxes on a frictionless surface are connected by a heavy cord of mass 1kg as shown in Fig. 4-53. If box A has a mass of 10kg, box B has a mass of 12kg, and FP = 40N, what is the acceleration of box B? Ignore sagging of the cord.
Hint: Don't overthink this! You can work this out conceptually just by using logic.
QUESTION 15
Suppose two boxes on a frictionless surface are connected by a heavy cord of mass 1kg as shown in Fig. 4-53. If box A has a mass of 10kg, box B has a mass of 12kg, and FP = 40N, what is the force, FBT, on the left box? Ignore sagging of the cord. 121.74
QUESTION 16
Suppose two boxes on a frictionless surface are connected by a heavy cord of mass 1kg as shown in Fig. 4-53. If box A has a mass of 10kg, box B has a mass of 12kg, and FP = 40N, what is the tension, FTA, on the right box? Ignore sagging of the cord. 20.8811.74

Explanation / Answer

10)

46.61 - uk * m * g = 5.57 x 0.68 {M=5.57}

so, uk = 0.783

11)

Providing that
Box mass = 12.21kg
Angle of incline = 31 degrees
Acceleration = 0.35 m/s/s

A)

First, the direction down the slope we will deem to be positive

Force down the slope without friction = mg sin()

F1= 12.21 x 9.81 x sin (31)

= 61.69 N

Net Force down the slope = ma

F2 = 12.21 x 0.35

= 4.2735 N

Net Force = F1 + F(friction)

F2 = F1 + Ff

4.2735- F1 = Ff

Ff = -57.4165 N

(negative since it is acting opposite F1 which we deemed positive)

B)

Ff = x N

57.4165 = x mg cos()

= 57.4165 / 154.52

= 0.56

Related Questions

Navigate

• Browse (All)
• Browse Q
• Subjects

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.