Answer:
The magnitude of the force exerted by the biceps [tex]f_b_i_c_e_p_s[/tex] is 218.38N
The magnitude of the force exerted by the elbow [tex]f_e_l_b_o_w[/tex] is 194.37N
Explanation:
Firstly , lets calculate the magnitude of force exerted by biceps [tex]f_b_i_c_e_p_s[/tex] -
Mass of the forearm=
[tex]m_f=1.50kg[/tex]
Mass of the ball =
[tex]m_b_a_l_l=950g[/tex]
= 0.950kg
so, weight of the forearm=
[tex]w_f=m_f[/tex] g
weight of the ball=
[tex]w_b_a_l_l=m_b_a_l_l[/tex] g
Now , balancing torque about the elbow , we have-
[tex]F_b_i_c_e_p_s\times d_b_i_c_e_p_s= w_f\times\frac{d_b_a_l_l}{2} + w_b_a_l_l\times d_b_a_l_l[/tex]
(here , distance of the bicep is 2.50cm = 0.025m and distance of the ball is 36cm= 0.36m)
Now putting the given values in the formula -
[tex]F_b_i_c_e_p_s\times 0.025= 0.95\times9.8\times\frac{0.36}{2} +1.50\times 0.36[/tex]
[tex]F_b_i_c_e_p_s=218.38N[/tex]
Now, lets calculate the force exerted by elbow [tex]F_e_l_b_o_w[/tex] -
Balancing the vertical forces
[tex]F_b_i_c_e_p_s=F_e_l_b_o_w+w_f+w_b_a_l_l[/tex]
[tex]F_e_l_b_o_w = 218.38-(1.00\times9.8)-(0.95\times9.8)[/tex]
[tex]F_e_l_b_o_w=194.37N[/tex]
