An elevator suspended by a vertical cable is moving downward but slowing down. The tension in the cable must be greater than the weight of the elevator. equal to the weight of the elevator. less than the weight of the elevator. O There is not enough information.
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- An elevator is supported by a cable and moving downward through the elevator shaft at a constant speed. How does the upward tension force compare to the downward force of gravity? O The upward tension force is stronger than the downward force of gravity. The downward force of gravity is stronger than the upward tension force. The tension force and the force of gravity are of equal strength.Problem 6: In the two cases shown below, a person is standing on a scale in an elevator. The elevators are identical, and the person weighs 500 N. In Case A the elevator is moving upward at a constant speed, and in Case B the elevator is moving downward at a constant speed. A в y = 3 m/s v = 3 m/s Will the scale reading in Case A be greater than, less than, or the same as the scale reading in Case B? Explain with a calculation or in words.E.) What is the tension at the botom of the rope? The force is upward. Neglect the air resistance F = 200 N 6.00 kg 4.00 kg 5.00 kg
- A rock is suspended from a string, and it accelerates upward. Which statement is true concerning the tension in the string? O The tension points downward. O The tension is less than the weight of the rock. O The tension is greater than the weight of the rock () The tension Is independent of the magnitude of the rocks acceleration, O The tension is equal to the weight of the rockTwo masses are connected by a rigid link as shown in Fig. Q3. The masses are resting on an inclined plane. The masses are then released. Considering the coefficient of kinetic friction between the inclined plane and the masses A and B are 0.1 and 0.2 respectively. 3. Compute by using Newton's second law a) The acceleration of both masses. b) The force in the rigid link and specify whether it is in tension or compression. 8kg В 4kg 25° Fig. Q3While digging a small hole prior to planting a tree, a homeowner encounters rocks. If he exerts a horizontal 239-N force on the prybar as shown, what is the horizontal force exerted on rock C? Note that a small ledge on rock C supports a vertical force reaction there. Neglect friction at B. Complete solutions (a) including and (b) excluding the weight of the 17-kg prybar. A 239 N 1550 mm 16° B 230 mm Answers: (a) Including weight, Fc- i (a) Excluding weight, Fe- i N
- Two packing crates of masses m, = 10.0 kg and m, = 7.30 kg are connected by a light string that passes over a frictionless pulley as in the figure below. The 7.30-kg crate lies on a smooth incline of angle 43.0°. Find the following. m, (a) the acceleration of the 7.30-kg crate m/s2 (up the incline) (b) the tension in the string NM Assume no friction M Hanging mass is 8 kg. Mass on table is 9 kg. What is the tension in the string connecting them? Answer to 2 sig figs.A traffic light weighing 122 N hangs from a cable tied to two other cablesfastened to a support. The upper cables make angles of 37.0◦ and 53.0◦with the horizontal. These upper cables are not as strong as the verticalcable, and will break if the tension in them exceeds 100 N. Will the trafficlight remain hanging in this situation, or will one of the cables break?
- (a) Calculate the tension (in N) in a vertical strand of spiderweb if a spider of mass 5.00 x 10¯º kg hangs motionless on it. (Enter a number.) N (b) Calculate the tension (in N) in a horizontal strand of spiderweb if the same spider sits motionless in the middle of it much like the tightrope walker in the figure. 5.0° 5.0° T. w T. TR The strand sags at an angle of 12.0° below the horizontal. (Enter a number.) N Compare this with the tension in the rtical strand (find their ratio). (Enter a number.) (tension in horizontal strand) / (tension in vertical strand)(a) Calculate the tension (in N) in a vertical strand of spiderweb if a spider of mass 8.00 x 10-5 kg hangs motionless on it. (Enter a number.) (b) Calculate the tension (in N) in a horizontal strand of spiderweb if the same spider sits motionless in the middle of it much like the tightrope walker in the figure. 5.0° 5.0° T. y4 T. Ta The strand sags at an angle of 12.0° below the horizontal. (Enter a number.) Compare this with the tension in the vertical strand (find their ratio). (Enter a number.) (tension in horizontal strand) / (tension in vertical strand) =Calculate the tension (in N) in a vertical strand of spiderweb if a spider of mass 5.00 x 10¯5 kg hangs motionless on it. Calculate the tension (in N) in a horizontal strand of spiderweb if the same spider sits motionless in the middle of it much like the tightrope walker 5.0° 5.00 T. TR y 4 T.I TR The strand sags at an angle of 11.0° below the horizontal. Compare this with the tension in the vertical strand (find their ratio). tension in horizontal strand tension in vertical strand