Question

The force F acts on the gripper of the robot arm. The moments of F about points A and B are 210N.m and 90N.m, respectively- both counterclockwise. determine F and the angle Q.​

Answer 1

Answer:

Explanation:this the question right

Answer 2

As a result,  the force F is 3 N and the angle Q is arccos(sqrt(1 - (70 / dA)^2)). The value of dA was not provided, so the exact value of Q cannot be determined.

• The principle of moments may be used to calculate the force F and the angle Q. Let dA signify the distance from point A to the location where force F is applied, and dB denote the distance from point B to the same location. Also, let Q represent the angle formed between the force and the line that connects points A and B.

• We are aware that the moment of the force F is 210 N.m. about point A and 90 N.m. around point B. Two equations can be established:

• Moment around point A is equal to F * dA * sin(Q) = 210 N.m.

• Moment around point B: 90 N.m = F * dB * sin(Q)

• When we divide the two equations, we obtain:

• F * dB / F * dA = 3 / 7 dB / 21 dB

• To solve for dB, we obtain:

• dB = (3 / 7) * dA

• This is what we get when we plug this into one of the moment equations:

• F*dA*sin(Q) = 210 N.m

• When we replace dB in terms of dA, we obtain:

• F*dA*sin(Q) = 210 N.m

• F = 90 N.m. when (3 / 7) * dA * sin(Q)

• When we divide the two equations, we obtain:

• F* (3/7)*dA / F*dA = 90 Nm / 210 Nm

• F = (7/3)*(90 N.m./210 N.m.) = 3 N

• We may use one of the moment equations to find sin(Q) now that we have determined F:

• F*dA*sin(Q) = 210 N.m

• Sin(Q) = 210 N.m / (3 N * dA) = 70 / dA. 3 N * dA * sin(Q) = 210 N.m

• Finally, we can calculate the angle Q using trigonometry:

• sqrt(1 - sin2(Q)) = cos(Q)

• sqrt(1 - (70 / dA)2) = cos(Q)

• Q is equal to arccos(sqrt(1 - (70/dA)2)).

For more questions on Science

https://brainly.in/question/24576823

#SPJ3