Question

In a study of the biomechanics of the tug-of-war, a 2.0-m-tall, 80.0-kg competitor in the middle of the line is considered to be a rigid body leaning back at an angle of 30.0? to the vertical. The competitor is pulling on a rope that is held horizontal a distance of 1.5m from his feet (as measured along the line of the body). At the moment shown in the (Figure 1) , the man is stationary and the tension in the rope in front of him is T1 = 1160N . Since there is friction between the rope and his hands, the tension in the rope behind him, T2, is not equal to T1. His center of mass is halfway between his feet and the top of his head. The coefficient of static friction between his feet and the ground is 0.65.

A) What is tension T2 in the rope behind him?

1)590 N

2)650 N      

3)860 N

4)1100 N

B) If he leans slightly farther back (increasing the angle between his body and the vertical) but remains stationary in this new position, which of the following statements is true? Assume that the rope remains horizontal.

1)The difference between T1 and T2 will increase, balancing the increased torque about his feet that his weight produces when he leans farther back.
2)The difference between T1 and T2 will decrease, balancing the increased torque about his feet that his weight produces when he leans farther back. 3)Neither T1 nor T2 will change, because no other forces are changing. 4)Both T1 and T2 will change, but the difference between them will remain the same.

C)What could explain this observation? As the holding height increases,

1)his center of mass moves down to compensate, so less tension in the rope is required to maintain equilibrium.
2) the moment arm of the rope about his feet decreases due to the angle that his body makes with the vertical. 3-) a smaller tension in the rope is needed to produce a torque sufficient to balance the torque of the weight about his feet. 4) the moment arm of the weight about his feet decreases due to the angle that his body makes with the vertical.

4) His body is leaning back at30.0 to the vertical, but the coefficient of static friction between his feet and the ground is suddenly reduced to 0.50. What will happen?

1) His feet will slip backward.

2) His feet will slip backward.

3) His feet will not slip.

4) His entire body will accelerate forward.

Answer 1

The following free body diagram shows that the forces acting on the body is, Tcos30 T, sin30° 30° 30° Tsin30° T, cos30° 30°

From the above figure, the weight is, W -mg -(80kg)(9.8m/$) -784J -(80kg) (9.8m/s') The net torque acting on point A is, T (d cos 30°)(dcos 30°)-W(dsin 30°) T. (1.5mcos 30°)-T (1.5mcos 30°)-mg(1.5msin 300) 1160N (1.5m cos30°)-784N (1.0m sin 30°) 1.5mcos 30° -858.21N 860N Therefore, option (3) is correct. If he leans slightly farther back, then the weight is increased and T is increased. So, the difference between T and T will increase. Therefore, option (1) is correct.

From the above figure the torque acting on the person is zero. T. (dcos 30°)-T (dcos 30°)-W (dsin 30°)-0 (T-T,)-r(sin 30) cos 30° The shortest distance between the rope and the ground is increased needed a smaller tension in the rope to produce a torque sufficient to balance the torque of the weight about his feet. Therefore, option (4) is corect. The forces acting along horizontal are zero. -1160N-858.5N 301.78N If the coefficient of static friction between his feet and the ground is suddenly reduced to 0.50 then the static frictional force is,

(0.5)(80kg) (9.8m/s) -(0.5) (80kg) (9.8m/s -392N To balancing the tension difference, he needs only 301.78 N. Even now the static friction force is greater as 392 N. So, his feet will not slip. Therefore, option (3) is coirect.