168,979 views
14 votes
14 votes
a. Find the sum of the horizontal forces on this system of wrenches. b. Find the sum of the vertical forces on this system of wrenches.c. If each of the forces is 4.0 cm from the center, find the sum of the torques about the center.

a. Find the sum of the horizontal forces on this system of wrenches. b. Find the sum-example-1
User John Flinchbaugh
by
3.0k points

2 Answers

18 votes
18 votes

Final answer:

The problem requires the calculation of the sum of horizontal and vertical forces and the net torque on a system. The steps involve using the torque formula and summing the individual torques with respect to an origin, taking into consideration each force's magnitude, direction, and point of application.

Step-by-step explanation:

The question relates to the concept of torque in physics, which is a measure of the turning force on an object that can cause it to rotate about an axis or pivot. Torque (T) is calculated using the formula T = rF sin θ, where r is the distance from the pivot point to the point where the force is applied, F is the magnitude of the force, and θ is the angle between the force vector and the arm of the lever. In scenarios involving equilibrium, the net force and net torque are zero (net F = 0 and net τ = 0 respectively).

For the given problem, one would start by identifying all the forces acting in the horizontal and vertical directions, drawing a free body diagram if needed, applying the right signs for each force depending on its direction, and then finding the sum of horizontal and vertical forces separately. Once the forces are summed up, the next step would be to calculate the torque for each force about the center using the provided distances and then summing all the individual torques to get the net torque about the center.

User Lazy Rabbit
by
2.6k points
29 votes
29 votes

ANSWERS

a. 0

b. 0

c. 400Nm

EXPLANATION

a. Two horizontal forces are acting on the system of wrenches: one is acting on the top wrench to the left and the other is acting on the bottom wrench to the right and both have the same magnitude. The sum is,


25N-25N=0N

b. Also, there are two vertical forces in opposite directions acting on the system of wrenches, on the left and right wrenches. Hence, the sum of the vertical forces is also 0

c. All the forces are acting on the wrenches in the counterclockwise direction so, by the right-hand rule, the torque made by each of these forces is in the same direction, from the center of the system of wrenches towards the outside of the screen. And, since all the forces are the same magnitude and are at the same distance from the center, they all produce the same torque,


\tau=25N*0.04m=100Nm

The sum is four times one of the torques,


\tau_(total)=4\cdot100Nm=400Nm

User Bvanderveen
by
2.9k points