The unknown tension and masses.

Question

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Answer 1

Question

Chapter 4, Problem 52P

(a)

To determine

The unknown tension and masses.

(a)

Expert Solution

Answer to Problem 52P

The value of the tension $T1$ is $60 N$ , $T2$ is $52 N$ and the value of the mass is $m$ is $5.3 kg$ .

Explanation of Solution

Given:

The given diagram is shown in Figure 1

PHYSICS F/SCI.+ENGRS.,STAND.-W/ACCESS, Chapter 4, Problem 52P , additional homework tip 1

Figure 1

Formula used:

For the given diagram, the expression for the horizontal forces is given by,

$T1cos60°=T3T1=T3cos60°$

The equation for the forces resolved in the vertical direction is given by,

$T2=T1sin60°$

The expression for the mass of the block is given by,

$m=T2g$

Calculation:

The tension $T1$ is calculated as,

$T1=T3cos60°=60 Ncos60°=60 N$

The expression for the value of the tension $T2$ in the rope is calculated as,

$T2=T1sin60°=(60 N)sin60°=52 N$

The mass of the block is calculated as,

$m=T2g=52 N9.8 m/ s 2=5.3 kg$

Conclusion:

Therefore, the value of the tension $T1$ is $60 N$ , $T2$ is $52 N$ and the value of the mass is $m$ is $5.3 kg$ .

(b)

To determine

The unknown tension and masses.

(b)

Expert Solution

Answer to Problem 52P

The value of the tension $T1$ is $42.6 N$ , $T2$ is $46.2 N$ and the value of the mass is $m$ is $4.7 kg$ .

Explanation of Solution

Formula used:

The free body diagram for the tension in the rope for figure (b) is shown in Figure 2

PHYSICS F/SCI.+ENGRS.,STAND.-W/ACCESS, Chapter 4, Problem 52P , additional homework tip 2

Figure 2

For the given diagram, the expression for the horizontal forces is given by,

$T1sin60°=T3sin60°T1=T3tan60°$

The equation for the forces resolved in the vertical direction is given by,

$T2=T3sin60°−T1cos60°$

The expression for the mass of the block is given by,

$m=T2g$

Calculation:

The tension $T1$ is calculated as,

$T1=T3tan60°=80 Ntan60°=42.6 N$

The expression for the value of the tension $T2$ in the rope is calculated as,

$T2=T3sin60°−T1cos60°=(80 N)sin60°−(42.6 N)cos60°=46.2 N$

The mass of the block is calculated as,

$m=T2g=46.2 N9.8 m/ s 2=4.7 kg$

Conclusion:

Therefore, the value of the tension $T1$ is $42.6 N$ , $T2$ is $46.2 N$ and the value of the mass is $m$ is $4.7 kg$ .

(c)

To determine

The unknown tension and masses.

(c)

Expert Solution

Answer to Problem 52P

The value of the tension $T1$ is $34 N$ , $T2$ is $58.8 N$ and the value of the mass is $m$ is $3.47 kg$ .

Explanation of Solution

Formula used:

The free body diagram for the tension in the rope for figure (c) is shown in Figure 3

PHYSICS F/SCI.+ENGRS.,STAND.-W/ACCESS, Chapter 4, Problem 52P , additional homework tip 3

Figure 3

For the given diagram, the expression for the horizontal forces is given by,

$T2=m1g$

The equation for the forces resolved in the vertical direction is given by,

$T2=T1sin60°+T3sin60°$

The expression for the mass of the block is given by,

$m=T1g$

The expression for the forces resolved in the horizontal direction is given by,

$T1cos60°=T3cos60°T1=T3$

Calculation:

The tension $T2$ is calculated as,

$T2=m1g=(6 kg)(9.8 m/ s 2)=58.8 N$

The expression for the value of the tension $T1$ in the rope is calculated as,

$T2=T1sin60°+T3sin60°58.8 N=T1sin60°+T1sinsin60°T1=58.8 N2sin60°T1=34 N$

The mass of the block is calculated as,

$m=T1g=34 N9.8 m/ s 2=3.47 kg$

Conclusion:

Therefore, the value of the tension $T1$ is $34 N$ , $T2$ is $58.8 N$ and the value of the mass is $m$ is $3.47 kg$ .

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Answer 2

Question

Chapter 4, Problem 52P

(a)

To determine

The unknown tension and masses.

(a)

Expert Solution

Answer to Problem 52P

The value of the tension $T1$ is $60 N$ , $T2$ is $52 N$ and the value of the mass is $m$ is $5.3 kg$ .

Explanation of Solution

Given:

The given diagram is shown in Figure 1

PHYSICS F/SCI.+ENGRS.,STAND.-W/ACCESS, Chapter 4, Problem 52P , additional homework tip 1

Figure 1

Formula used:

For the given diagram, the expression for the horizontal forces is given by,

$T1cos60°=T3T1=T3cos60°$

The equation for the forces resolved in the vertical direction is given by,

$T2=T1sin60°$

The expression for the mass of the block is given by,

$m=T2g$

Calculation:

The tension $T1$ is calculated as,

$T1=T3cos60°=60 Ncos60°=60 N$

The expression for the value of the tension $T2$ in the rope is calculated as,

$T2=T1sin60°=(60 N)sin60°=52 N$

The mass of the block is calculated as,

$m=T2g=52 N9.8 m/ s 2=5.3 kg$

Conclusion:

Therefore, the value of the tension $T1$ is $60 N$ , $T2$ is $52 N$ and the value of the mass is $m$ is $5.3 kg$ .

(b)

To determine

The unknown tension and masses.

(b)

Expert Solution

Answer to Problem 52P

The value of the tension $T1$ is $42.6 N$ , $T2$ is $46.2 N$ and the value of the mass is $m$ is $4.7 kg$ .

Explanation of Solution

Formula used:

The free body diagram for the tension in the rope for figure (b) is shown in Figure 2

PHYSICS F/SCI.+ENGRS.,STAND.-W/ACCESS, Chapter 4, Problem 52P , additional homework tip 2

Figure 2

For the given diagram, the expression for the horizontal forces is given by,

$T1sin60°=T3sin60°T1=T3tan60°$

The equation for the forces resolved in the vertical direction is given by,

$T2=T3sin60°−T1cos60°$

The expression for the mass of the block is given by,

$m=T2g$

Calculation:

The tension $T1$ is calculated as,

$T1=T3tan60°=80 Ntan60°=42.6 N$

The expression for the value of the tension $T2$ in the rope is calculated as,

$T2=T3sin60°−T1cos60°=(80 N)sin60°−(42.6 N)cos60°=46.2 N$

The mass of the block is calculated as,

$m=T2g=46.2 N9.8 m/ s 2=4.7 kg$

Conclusion:

Therefore, the value of the tension $T1$ is $42.6 N$ , $T2$ is $46.2 N$ and the value of the mass is $m$ is $4.7 kg$ .

(c)

To determine

The unknown tension and masses.

(c)

Expert Solution

Answer to Problem 52P

The value of the tension $T1$ is $34 N$ , $T2$ is $58.8 N$ and the value of the mass is $m$ is $3.47 kg$ .

Explanation of Solution

Formula used:

The free body diagram for the tension in the rope for figure (c) is shown in Figure 3

PHYSICS F/SCI.+ENGRS.,STAND.-W/ACCESS, Chapter 4, Problem 52P , additional homework tip 3

Figure 3

For the given diagram, the expression for the horizontal forces is given by,

$T2=m1g$

The equation for the forces resolved in the vertical direction is given by,

$T2=T1sin60°+T3sin60°$

The expression for the mass of the block is given by,

$m=T1g$

The expression for the forces resolved in the horizontal direction is given by,

$T1cos60°=T3cos60°T1=T3$

Calculation:

The tension $T2$ is calculated as,

$T2=m1g=(6 kg)(9.8 m/ s 2)=58.8 N$

The expression for the value of the tension $T1$ in the rope is calculated as,

$T2=T1sin60°+T3sin60°58.8 N=T1sin60°+T1sinsin60°T1=58.8 N2sin60°T1=34 N$

The mass of the block is calculated as,

$m=T1g=34 N9.8 m/ s 2=3.47 kg$

Conclusion:

Therefore, the value of the tension $T1$ is $34 N$ , $T2$ is $58.8 N$ and the value of the mass is $m$ is $3.47 kg$ .

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Answer 3

Question

Chapter 4, Problem 52P

(a)

To determine

The unknown tension and masses.

(a)

Expert Solution

Answer to Problem 52P

The value of the tension $T1$ is $60 N$ , $T2$ is $52 N$ and the value of the mass is $m$ is $5.3 kg$ .

Explanation of Solution

Given:

The given diagram is shown in Figure 1

PHYSICS F/SCI.+ENGRS.,STAND.-W/ACCESS, Chapter 4, Problem 52P , additional homework tip 1

Figure 1

Formula used:

For the given diagram, the expression for the horizontal forces is given by,

$T1cos60°=T3T1=T3cos60°$

The equation for the forces resolved in the vertical direction is given by,

$T2=T1sin60°$

The expression for the mass of the block is given by,

$m=T2g$

Calculation:

The tension $T1$ is calculated as,

$T1=T3cos60°=60 Ncos60°=60 N$

The expression for the value of the tension $T2$ in the rope is calculated as,

$T2=T1sin60°=(60 N)sin60°=52 N$

The mass of the block is calculated as,

$m=T2g=52 N9.8 m/ s 2=5.3 kg$

Conclusion:

Therefore, the value of the tension $T1$ is $60 N$ , $T2$ is $52 N$ and the value of the mass is $m$ is $5.3 kg$ .

(b)

To determine

The unknown tension and masses.

(b)

Expert Solution

Answer to Problem 52P

The value of the tension $T1$ is $42.6 N$ , $T2$ is $46.2 N$ and the value of the mass is $m$ is $4.7 kg$ .

Explanation of Solution

Formula used:

The free body diagram for the tension in the rope for figure (b) is shown in Figure 2

PHYSICS F/SCI.+ENGRS.,STAND.-W/ACCESS, Chapter 4, Problem 52P , additional homework tip 2

Figure 2

For the given diagram, the expression for the horizontal forces is given by,

$T1sin60°=T3sin60°T1=T3tan60°$

The equation for the forces resolved in the vertical direction is given by,

$T2=T3sin60°−T1cos60°$

The expression for the mass of the block is given by,

$m=T2g$

Calculation:

The tension $T1$ is calculated as,

$T1=T3tan60°=80 Ntan60°=42.6 N$

The expression for the value of the tension $T2$ in the rope is calculated as,

$T2=T3sin60°−T1cos60°=(80 N)sin60°−(42.6 N)cos60°=46.2 N$

The mass of the block is calculated as,

$m=T2g=46.2 N9.8 m/ s 2=4.7 kg$

Conclusion:

Therefore, the value of the tension $T1$ is $42.6 N$ , $T2$ is $46.2 N$ and the value of the mass is $m$ is $4.7 kg$ .

(c)

To determine

The unknown tension and masses.

(c)

Expert Solution

Answer to Problem 52P

The value of the tension $T1$ is $34 N$ , $T2$ is $58.8 N$ and the value of the mass is $m$ is $3.47 kg$ .

Explanation of Solution

Formula used:

The free body diagram for the tension in the rope for figure (c) is shown in Figure 3

PHYSICS F/SCI.+ENGRS.,STAND.-W/ACCESS, Chapter 4, Problem 52P , additional homework tip 3

Figure 3

For the given diagram, the expression for the horizontal forces is given by,

$T2=m1g$

The equation for the forces resolved in the vertical direction is given by,

$T2=T1sin60°+T3sin60°$

The expression for the mass of the block is given by,

$m=T1g$

The expression for the forces resolved in the horizontal direction is given by,

$T1cos60°=T3cos60°T1=T3$

Calculation:

The tension $T2$ is calculated as,

$T2=m1g=(6 kg)(9.8 m/ s 2)=58.8 N$

The expression for the value of the tension $T1$ in the rope is calculated as,

$T2=T1sin60°+T3sin60°58.8 N=T1sin60°+T1sinsin60°T1=58.8 N2sin60°T1=34 N$

The mass of the block is calculated as,

$m=T1g=34 N9.8 m/ s 2=3.47 kg$

Conclusion:

Therefore, the value of the tension $T1$ is $34 N$ , $T2$ is $58.8 N$ and the value of the mass is $m$ is $3.47 kg$ .

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Answer 4

Question

Chapter 4, Problem 52P

(a)

To determine

The unknown tension and masses.

(a)

Expert Solution

Answer to Problem 52P

The value of the tension $T1$ is $60 N$ , $T2$ is $52 N$ and the value of the mass is $m$ is $5.3 kg$ .

Explanation of Solution

Given:

The given diagram is shown in Figure 1

PHYSICS F/SCI.+ENGRS.,STAND.-W/ACCESS, Chapter 4, Problem 52P , additional homework tip 1

Figure 1

Formula used:

For the given diagram, the expression for the horizontal forces is given by,

$T1cos60°=T3T1=T3cos60°$

The equation for the forces resolved in the vertical direction is given by,

$T2=T1sin60°$

The expression for the mass of the block is given by,

$m=T2g$

Calculation:

The tension $T1$ is calculated as,

$T1=T3cos60°=60 Ncos60°=60 N$

The expression for the value of the tension $T2$ in the rope is calculated as,

$T2=T1sin60°=(60 N)sin60°=52 N$

The mass of the block is calculated as,

$m=T2g=52 N9.8 m/ s 2=5.3 kg$

Conclusion:

Therefore, the value of the tension $T1$ is $60 N$ , $T2$ is $52 N$ and the value of the mass is $m$ is $5.3 kg$ .

(b)

To determine

The unknown tension and masses.

(b)

Expert Solution

Answer to Problem 52P

The value of the tension $T1$ is $42.6 N$ , $T2$ is $46.2 N$ and the value of the mass is $m$ is $4.7 kg$ .

Explanation of Solution

Formula used:

The free body diagram for the tension in the rope for figure (b) is shown in Figure 2

PHYSICS F/SCI.+ENGRS.,STAND.-W/ACCESS, Chapter 4, Problem 52P , additional homework tip 2

Figure 2

For the given diagram, the expression for the horizontal forces is given by,

$T1sin60°=T3sin60°T1=T3tan60°$

The equation for the forces resolved in the vertical direction is given by,

$T2=T3sin60°−T1cos60°$

The expression for the mass of the block is given by,

$m=T2g$

Calculation:

The tension $T1$ is calculated as,

$T1=T3tan60°=80 Ntan60°=42.6 N$

The expression for the value of the tension $T2$ in the rope is calculated as,

$T2=T3sin60°−T1cos60°=(80 N)sin60°−(42.6 N)cos60°=46.2 N$

The mass of the block is calculated as,

$m=T2g=46.2 N9.8 m/ s 2=4.7 kg$

Conclusion:

Therefore, the value of the tension $T1$ is $42.6 N$ , $T2$ is $46.2 N$ and the value of the mass is $m$ is $4.7 kg$ .

(c)

To determine

The unknown tension and masses.

(c)

Expert Solution

Answer to Problem 52P

The value of the tension $T1$ is $34 N$ , $T2$ is $58.8 N$ and the value of the mass is $m$ is $3.47 kg$ .

Explanation of Solution

Formula used:

The free body diagram for the tension in the rope for figure (c) is shown in Figure 3

PHYSICS F/SCI.+ENGRS.,STAND.-W/ACCESS, Chapter 4, Problem 52P , additional homework tip 3

Figure 3

For the given diagram, the expression for the horizontal forces is given by,

$T2=m1g$

The equation for the forces resolved in the vertical direction is given by,

$T2=T1sin60°+T3sin60°$

The expression for the mass of the block is given by,

$m=T1g$

The expression for the forces resolved in the horizontal direction is given by,

$T1cos60°=T3cos60°T1=T3$

Calculation:

The tension $T2$ is calculated as,

$T2=m1g=(6 kg)(9.8 m/ s 2)=58.8 N$

The expression for the value of the tension $T1$ in the rope is calculated as,

$T2=T1sin60°+T3sin60°58.8 N=T1sin60°+T1sinsin60°T1=58.8 N2sin60°T1=34 N$

The mass of the block is calculated as,

$m=T1g=34 N9.8 m/ s 2=3.47 kg$

Conclusion:

Therefore, the value of the tension $T1$ is $34 N$ , $T2$ is $58.8 N$ and the value of the mass is $m$ is $3.47 kg$ .

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Answer 5

Chapter 4, Problem 52P

(a)

To determine

The unknown tension and masses.

(a)

Expert Solution

Answer to Problem 52P

The value of the tension $T1$ is $60 N$ , $T2$ is $52 N$ and the value of the mass is $m$ is $5.3 kg$ .

Explanation of Solution

Given:

The given diagram is shown in Figure 1

PHYSICS F/SCI.+ENGRS.,STAND.-W/ACCESS, Chapter 4, Problem 52P , additional homework tip 1

Figure 1

Formula used:

For the given diagram, the expression for the horizontal forces is given by,

$T1cos60°=T3T1=T3cos60°$

The equation for the forces resolved in the vertical direction is given by,

$T2=T1sin60°$

The expression for the mass of the block is given by,

$m=T2g$

Calculation:

The tension $T1$ is calculated as,

$T1=T3cos60°=60 Ncos60°=60 N$

The expression for the value of the tension $T2$ in the rope is calculated as,

$T2=T1sin60°=(60 N)sin60°=52 N$

The mass of the block is calculated as,

$m=T2g=52 N9.8 m/ s 2=5.3 kg$

Conclusion:

Therefore, the value of the tension $T1$ is $60 N$ , $T2$ is $52 N$ and the value of the mass is $m$ is $5.3 kg$ .

(b)

To determine

The unknown tension and masses.

(b)

Expert Solution

Answer to Problem 52P

The value of the tension $T1$ is $42.6 N$ , $T2$ is $46.2 N$ and the value of the mass is $m$ is $4.7 kg$ .

Explanation of Solution

Formula used:

The free body diagram for the tension in the rope for figure (b) is shown in Figure 2

PHYSICS F/SCI.+ENGRS.,STAND.-W/ACCESS, Chapter 4, Problem 52P , additional homework tip 2

Figure 2

For the given diagram, the expression for the horizontal forces is given by,

$T1sin60°=T3sin60°T1=T3tan60°$

The equation for the forces resolved in the vertical direction is given by,

$T2=T3sin60°−T1cos60°$

The expression for the mass of the block is given by,

$m=T2g$

Calculation:

The tension $T1$ is calculated as,

$T1=T3tan60°=80 Ntan60°=42.6 N$

The expression for the value of the tension $T2$ in the rope is calculated as,

$T2=T3sin60°−T1cos60°=(80 N)sin60°−(42.6 N)cos60°=46.2 N$

The mass of the block is calculated as,

$m=T2g=46.2 N9.8 m/ s 2=4.7 kg$

Conclusion:

Therefore, the value of the tension $T1$ is $42.6 N$ , $T2$ is $46.2 N$ and the value of the mass is $m$ is $4.7 kg$ .

(c)

To determine

The unknown tension and masses.

(c)

Expert Solution

Answer to Problem 52P

The value of the tension $T1$ is $34 N$ , $T2$ is $58.8 N$ and the value of the mass is $m$ is $3.47 kg$ .

Explanation of Solution

Formula used:

The free body diagram for the tension in the rope for figure (c) is shown in Figure 3

PHYSICS F/SCI.+ENGRS.,STAND.-W/ACCESS, Chapter 4, Problem 52P , additional homework tip 3

Figure 3

For the given diagram, the expression for the horizontal forces is given by,

$T2=m1g$

The equation for the forces resolved in the vertical direction is given by,

$T2=T1sin60°+T3sin60°$

The expression for the mass of the block is given by,

$m=T1g$

The expression for the forces resolved in the horizontal direction is given by,

$T1cos60°=T3cos60°T1=T3$

Calculation:

The tension $T2$ is calculated as,

$T2=m1g=(6 kg)(9.8 m/ s 2)=58.8 N$

The expression for the value of the tension $T1$ in the rope is calculated as,

$T2=T1sin60°+T3sin60°58.8 N=T1sin60°+T1sinsin60°T1=58.8 N2sin60°T1=34 N$

The mass of the block is calculated as,

$m=T1g=34 N9.8 m/ s 2=3.47 kg$

Conclusion:

Therefore, the value of the tension $T1$ is $34 N$ , $T2$ is $58.8 N$ and the value of the mass is $m$ is $3.47 kg$ .

Answer 6

Chapter 4, Problem 52P

(a)

To determine

The unknown tension and masses.

(a)

Expert Solution

Answer to Problem 52P

The value of the tension $T1$ is $60 N$ , $T2$ is $52 N$ and the value of the mass is $m$ is $5.3 kg$ .

Explanation of Solution

Given:

The given diagram is shown in Figure 1

PHYSICS F/SCI.+ENGRS.,STAND.-W/ACCESS, Chapter 4, Problem 52P , additional homework tip 1

Figure 1

Formula used:

For the given diagram, the expression for the horizontal forces is given by,

$T1cos60°=T3T1=T3cos60°$

The equation for the forces resolved in the vertical direction is given by,

$T2=T1sin60°$

The expression for the mass of the block is given by,

$m=T2g$

Calculation:

The tension $T1$ is calculated as,

$T1=T3cos60°=60 Ncos60°=60 N$

The expression for the value of the tension $T2$ in the rope is calculated as,

$T2=T1sin60°=(60 N)sin60°=52 N$

The mass of the block is calculated as,

$m=T2g=52 N9.8 m/ s 2=5.3 kg$

Conclusion:

Therefore, the value of the tension $T1$ is $60 N$ , $T2$ is $52 N$ and the value of the mass is $m$ is $5.3 kg$ .

Answer 7

Chapter 4, Problem 52P

(a)

To determine

The unknown tension and masses.

Answer 8

(a)

Expert Solution

Answer to Problem 52P

The value of the tension $T1$ is $60 N$ , $T2$ is $52 N$ and the value of the mass is $m$ is $5.3 kg$ .

Answer 9

(a)

Expert Solution

Answer 10

(a)

Expert Solution

Answer 11

Expert Solution

Answer 12

Explanation of Solution

Given:

The given diagram is shown in Figure 1

PHYSICS F/SCI.+ENGRS.,STAND.-W/ACCESS, Chapter 4, Problem 52P , additional homework tip 1

Figure 1

Formula used:

For the given diagram, the expression for the horizontal forces is given by,

$T1cos60°=T3T1=T3cos60°$

The equation for the forces resolved in the vertical direction is given by,

$T2=T1sin60°$

The expression for the mass of the block is given by,

$m=T2g$

Calculation:

The tension $T1$ is calculated as,

$T1=T3cos60°=60 Ncos60°=60 N$

The expression for the value of the tension $T2$ in the rope is calculated as,

$T2=T1sin60°=(60 N)sin60°=52 N$

The mass of the block is calculated as,

$m=T2g=52 N9.8 m/ s 2=5.3 kg$

Conclusion:

Therefore, the value of the tension $T1$ is $60 N$ , $T2$ is $52 N$ and the value of the mass is $m$ is $5.3 kg$ .

Answer 13

(b)

To determine

The unknown tension and masses.

(b)

Expert Solution

Answer to Problem 52P

The value of the tension $T1$ is $42.6 N$ , $T2$ is $46.2 N$ and the value of the mass is $m$ is $4.7 kg$ .

Explanation of Solution

Formula used:

The free body diagram for the tension in the rope for figure (b) is shown in Figure 2

PHYSICS F/SCI.+ENGRS.,STAND.-W/ACCESS, Chapter 4, Problem 52P , additional homework tip 2

Figure 2

For the given diagram, the expression for the horizontal forces is given by,

$T1sin60°=T3sin60°T1=T3tan60°$

The equation for the forces resolved in the vertical direction is given by,

$T2=T3sin60°−T1cos60°$

The expression for the mass of the block is given by,

$m=T2g$

Calculation:

The tension $T1$ is calculated as,

$T1=T3tan60°=80 Ntan60°=42.6 N$

The expression for the value of the tension $T2$ in the rope is calculated as,

$T2=T3sin60°−T1cos60°=(80 N)sin60°−(42.6 N)cos60°=46.2 N$

The mass of the block is calculated as,

$m=T2g=46.2 N9.8 m/ s 2=4.7 kg$

Conclusion:

Therefore, the value of the tension $T1$ is $42.6 N$ , $T2$ is $46.2 N$ and the value of the mass is $m$ is $4.7 kg$ .

Answer 14

(b)

To determine

The unknown tension and masses.

Answer 15

(b)

Expert Solution

Answer to Problem 52P

The value of the tension $T1$ is $42.6 N$ , $T2$ is $46.2 N$ and the value of the mass is $m$ is $4.7 kg$ .

Answer 16

(b)

Expert Solution

Answer 17

(b)

Expert Solution

Answer 18

Expert Solution

Answer 19

Explanation of Solution

Formula used:

The free body diagram for the tension in the rope for figure (b) is shown in Figure 2

PHYSICS F/SCI.+ENGRS.,STAND.-W/ACCESS, Chapter 4, Problem 52P , additional homework tip 2

Figure 2

For the given diagram, the expression for the horizontal forces is given by,

$T1sin60°=T3sin60°T1=T3tan60°$

The equation for the forces resolved in the vertical direction is given by,

$T2=T3sin60°−T1cos60°$

The expression for the mass of the block is given by,

$m=T2g$

Calculation:

The tension $T1$ is calculated as,

$T1=T3tan60°=80 Ntan60°=42.6 N$

The expression for the value of the tension $T2$ in the rope is calculated as,

$T2=T3sin60°−T1cos60°=(80 N)sin60°−(42.6 N)cos60°=46.2 N$

The mass of the block is calculated as,

$m=T2g=46.2 N9.8 m/ s 2=4.7 kg$

Conclusion:

Therefore, the value of the tension $T1$ is $42.6 N$ , $T2$ is $46.2 N$ and the value of the mass is $m$ is $4.7 kg$ .

Answer 20

(c)

To determine

The unknown tension and masses.

(c)

Expert Solution

Answer to Problem 52P

The value of the tension $T1$ is $34 N$ , $T2$ is $58.8 N$ and the value of the mass is $m$ is $3.47 kg$ .

Explanation of Solution

Formula used:

The free body diagram for the tension in the rope for figure (c) is shown in Figure 3

PHYSICS F/SCI.+ENGRS.,STAND.-W/ACCESS, Chapter 4, Problem 52P , additional homework tip 3

Figure 3

For the given diagram, the expression for the horizontal forces is given by,

$T2=m1g$

The equation for the forces resolved in the vertical direction is given by,

$T2=T1sin60°+T3sin60°$

The expression for the mass of the block is given by,

$m=T1g$

The expression for the forces resolved in the horizontal direction is given by,

$T1cos60°=T3cos60°T1=T3$

Calculation:

The tension $T2$ is calculated as,

$T2=m1g=(6 kg)(9.8 m/ s 2)=58.8 N$

The expression for the value of the tension $T1$ in the rope is calculated as,

$T2=T1sin60°+T3sin60°58.8 N=T1sin60°+T1sinsin60°T1=58.8 N2sin60°T1=34 N$

The mass of the block is calculated as,

$m=T1g=34 N9.8 m/ s 2=3.47 kg$

Conclusion:

Therefore, the value of the tension $T1$ is $34 N$ , $T2$ is $58.8 N$ and the value of the mass is $m$ is $3.47 kg$ .

Answer 21

(c)

To determine

The unknown tension and masses.

Answer 22

(c)

Expert Solution

Answer to Problem 52P

The value of the tension $T1$ is $34 N$ , $T2$ is $58.8 N$ and the value of the mass is $m$ is $3.47 kg$ .

Answer 23

(c)

Expert Solution

Answer 24

(c)

Expert Solution

Answer 25

Expert Solution

Answer 26

Explanation of Solution

Formula used:

The free body diagram for the tension in the rope for figure (c) is shown in Figure 3

PHYSICS F/SCI.+ENGRS.,STAND.-W/ACCESS, Chapter 4, Problem 52P , additional homework tip 3

Figure 3

For the given diagram, the expression for the horizontal forces is given by,

$T2=m1g$

The equation for the forces resolved in the vertical direction is given by,

$T2=T1sin60°+T3sin60°$

The expression for the mass of the block is given by,

$m=T1g$

The expression for the forces resolved in the horizontal direction is given by,

$T1cos60°=T3cos60°T1=T3$

Calculation:

The tension $T2$ is calculated as,

$T2=m1g=(6 kg)(9.8 m/ s 2)=58.8 N$

The expression for the value of the tension $T1$ in the rope is calculated as,

$T2=T1sin60°+T3sin60°58.8 N=T1sin60°+T1sinsin60°T1=58.8 N2sin60°T1=34 N$

The mass of the block is calculated as,

$m=T1g=34 N9.8 m/ s 2=3.47 kg$

Conclusion:

Therefore, the value of the tension $T1$ is $34 N$ , $T2$ is $58.8 N$ and the value of the mass is $m$ is $3.47 kg$ .

Answer 27

Ch. 4 - Prob. 1P Ch. 4 - Prob. 2P Ch. 4 - Prob. 3P Ch. 4 - Prob. 4P Ch. 4 - Prob. 5P Ch. 4 - Prob. 6P Ch. 4 - Prob. 7P Ch. 4 - Prob. 8P Ch. 4 - Prob. 9P Ch. 4 - Prob. 10P

The unknown tension and masses.

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The unknown tension and masses.

Answer to Problem 52P

Explanation of Solution

The unknown tension and masses.

Answer to Problem 52P

Explanation of Solution

The unknown tension and masses.

Answer to Problem 52P

Explanation of Solution

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